Construction of Bhaskara Pairs

TL;DR

This paper constructs specific integer solutions for a system of coupled quadratic and cubic Diophantine equations with fixed ratios, advancing the understanding of such complex number solutions.

Contribution

It introduces a method to explicitly construct integer solutions to the coupled equations for given ratios a/b, which was previously unexplored.

Findings

Explicit solutions for specific ratios a/b

New insights into coupled quadratic-cubic Diophantine systems

Potential applications in number theory research

Abstract

We construct integer solutions {a,b} to the coupled system of diophantine quadratic-cubic equations a^2+b^2=x^3 and a^3+b^3=y^2 for fixed ratios a/b.

Tables

Table 1. Table 1. Prime factorizations of 1 + k 2 1 superscript 𝑘 2 1+k^{2} , 1 + k 3 1 superscript 𝑘 3 1+k^{3} and k 𝑘 k

$k$	$1+k^2$	$1+k^3$	$k$
1	2	2	1
2	5	$3^2$	2
3	$2\times 5$	$2^2\times 7$	3
4	17	$5\times 13$	$2^2$
5	$2\times 13$	$2\times 3^2\times 7$	$5$
6	$37$	$7\times 31$	$2\times 3$

Table 2. Table 2. Solutions b i − k i subscript 𝑏 𝑖 subscript 𝑘 𝑖 b_{i}-k_{i} to ( 18 ) as a function of c i ( mod 3 ) annotated subscript 𝑐 𝑖 pmod 3 c_{i}\pmod{3} and d i ( mod 2 ) annotated subscript 𝑑 𝑖 pmod 2 d_{i}\pmod{2} .

$c_i\backslash d_i$	0	1
0	0	3
1	4	1
2	2	5

Table 3. Table 3. The Chinese remainder solutions for u / k = 1 𝑢 𝑘 1 u/k=1 . Fundamental solution b = 2 𝑏 2 b=2 , a = 2 𝑎 2 a=2 .

$i$	$c_i$	$d_i$	$k_i$	$p_i^{b_i}$
1	1	1	0	$2^1$

Table 4. Table 4. The Chinese remainder solutions for u / k = 1 / 2 𝑢 𝑘 1 2 u/k=1/2 . Fundamental solution b = 2 × 5 4 𝑏 2 superscript 5 4 b=2\times 5^{4} , a = 5 4 𝑎 superscript 5 4 a=5^{4} .

$i$	$c_i$	$d_i$	$k_i$	$p_i^{b_i}$
1	0	0	1	$2^1$
2	0	2	0	$3^0$
3	1	0	0	$5^4$

Table 5. Table 5. The Chinese remainder solutions for u / k = 1 / 3 𝑢 𝑘 1 3 u/k=1/3 . Fundamental solution b = 2 4 × 3 × 5 4 × 7 3 𝑏 superscript 2 4 3 superscript 5 4 superscript 7 3 b=2^{4}\times 3\times 5^{4}\times 7^{3} , a = 2 4 × 5 4 × 7 3 𝑎 superscript 2 4 superscript 5 4 superscript 7 3 a=2^{4}\times 5^{4}\times 7^{3} .

$i$	$c_i$	$d_i$	$k_i$	$p_i^{b_i}$
1	1	2	0	$2^4$
2	0	0	1	$3^1$
3	1	0	0	$5^4$
4	0	1	0	$7^3$

Table 6. Table 6. The Chinese remainder solutions for u / k = 1 / 4 𝑢 𝑘 1 4 u/k=1/4 . Fundamental solution b = 2 2 × 5 3 × 13 3 × 17 4 𝑏 superscript 2 2 superscript 5 3 superscript 13 3 superscript 17 4 b=2^{2}\times 5^{3}\times 13^{3}\times 17^{4} , a = 5 3 × 13 3 × 17 4 𝑎 superscript 5 3 superscript 13 3 superscript 17 4 a=5^{3}\times 13^{3}\times 17^{4} .

$i$	$c_i$	$d_i$	$k_i$	$p_i^{b_i}$
1	0	0	2	$2^2$
3	0	1	0	$5^3$
6	0	1	0	${13}^3$
7	1	0	0	${17}^4$

Table 7. Table 7. The Chinese remainder solutions for u / k = 1 / 5 𝑢 𝑘 1 5 u/k=1/5 . Fundamental solution b = 2 × 5 × 7 3 × 13 4 𝑏 2 5 superscript 7 3 superscript 13 4 b=2\times 5\times 7^{3}\times 13^{4} , a = 2 × 7 3 × 13 4 𝑎 2 superscript 7 3 superscript 13 4 a=2\times 7^{3}\times 13^{4} .

$i$	$c_i$	$d_i$	$k_i$	$p_i^{b_i}$
1	1	1	0	$2^1$
2	0	2	0	$3^0$
3	0	0	1	$5^1$
4	0	1	0	$7^3$
6	1	0	0	${13}^4$

Table 8. Table 8. The Chinese remainder solutions for u / k = 1 / 6 𝑢 𝑘 1 6 u/k=1/6 . Fundamental solution b = 2 × 3 × 7 3 × 31 3 × 37 4 𝑏 2 3 superscript 7 3 superscript 31 3 superscript 37 4 b=2\times 3\times 7^{3}\times 31^{3}\times 37^{4} , a = 7 3 × 31 3 × 37 4 𝑎 superscript 7 3 superscript 31 3 superscript 37 4 a=7^{3}\times 31^{3}\times 37^{4} .

$i$	$c_i$	$d_i$	$k_i$	$p_i^{b_i}$
1	0	0	1	$2^1$
2	0	0	1	$3^1$
4	0	1	0	$7^3$
11	0	1	0	${31}^3$
12	1	0	0	${37}^4$

Table 9. Table 9. The Chinese remainder solutions for u / k = 3 / 4 𝑢 𝑘 3 4 u/k=3/4 . Fundamental solution b = 2 2 × 5 2 × 7 3 × 13 3 𝑏 superscript 2 2 superscript 5 2 superscript 7 3 superscript 13 3 b=2^{2}\times 5^{2}\times 7^{3}\times 13^{3} , a = 3 × 5 2 × 7 3 × 13 3 𝑎 3 superscript 5 2 superscript 7 3 superscript 13 3 a=3\times 5^{2}\times 7^{3}\times 13^{3} .

$i$	$c_i$	$d_i$	$k_i$	$p_i^{b_i}$
1	0	0	2	$2^2$
3	2	0	0	$5^2$
4	0	1	0	$7^3$
6	0	1	0	${13}^3$

Table 10. Table 10. The Chinese remainder solutions for u / k = 5 / 6 𝑢 𝑘 5 6 u/k=5/6 . Fundamental solution b = 2 × 3 × 11 3 × 31 3 × 61 4 𝑏 2 3 superscript 11 3 superscript 31 3 superscript 61 4 b=2\times 3\times 11^{3}\times 31^{3}\times 61^{4} , a = 5 × 11 3 × 31 3 × 61 4 𝑎 5 superscript 11 3 superscript 31 3 superscript 61 4 a=5\times 11^{3}\times 31^{3}\times 61^{4} .

$i$	$c_i$	$d_i$	$k_i$	$p_i^{b_i}$
1	0	0	1	$2^1$
2	0	0	1	$3^1$
5	0	1	0	${11}^3$
11	0	1	0	${31}^3$
18	1	0	0	${61}^4$

Table 11. Table 11. The Chinese remainder solutions for u / k = 2 / 11 𝑢 𝑘 2 11 u/k=2/11 , u 2 + k 2 = 5 3 superscript 𝑢 2 superscript 𝑘 2 superscript 5 3 u^{2}+k^{2}=5^{3} , u 3 + k 3 = 13 × 103 superscript 𝑢 3 superscript 𝑘 3 13 103 u^{3}+k^{3}=13\times 103 . Fundamental solution b = 11 × 13 3 × 103 3 𝑏 11 superscript 13 3 superscript 103 3 b=11\times 13^{3}\times 103^{3} , a = 2 × 13 3 × 103 3 𝑎 2 superscript 13 3 superscript 103 3 a=2\times 13^{3}\times 103^{3} .

$i$	$c_i$	$d_i$	$k_i$	$p_i^{b_i}$
3	3	0	0	$5^0$
5	0	0	1	${11}^1$
6	0	1	0	${13}^3$
27	0	1	0	${103}^3$

Table 12. Table 12. The Chinese remainder solutions for u / k = 7 / 17 𝑢 𝑘 7 17 u/k=7/17 . Fundamental solution b = 2 × 13 2 × 17 × 73 3 𝑏 2 superscript 13 2 17 superscript 73 3 b=2\times 13^{2}\times 17\times 73^{3} , a = 2 × 7 × 13 2 × 73 3 𝑎 2 7 superscript 13 2 superscript 73 3 a=2\times 7\times 13^{2}\times 73^{3} .

$i$	$c_i$	$d_i$	$k_i$	$p_i^{b_i}$
1	1	3	0	$2^1$
2	0	2	0	$3^0$
6	2	0	0	${13}^2$
7	0	0	1	${17}^1$
21	0	1	0	${73}^3$

Table 13. Table 13. The Chinese remainder solutions for u / k = 2 6 / 83 𝑢 𝑘 superscript 2 6 83 u/k=2^{6}/83 , Example 1 .

$i$	$c_i$	$d_i$	$k_i$	$p_i^{b_i}$
2	0	2	0	$3^0$
3	1	0	0	$5^4$
4	0	2	0	$7^0$
6	3	0	0	${13}^0$
11	0	1	0	${31}^3$
18	0	1	0	${61}^3$
23	0	0	1	${83}^1$

Table 14. Table 14. The fundamental solutions for ratios a / b = u / k 𝑎 𝑏 𝑢 𝑘 a/b=u/k up to denominator k = 11 𝑘 11 k=11 .

$a$	$b$	$u/k$
2	2	1
625	1250	1/2
3430000	10290000	1/3
2449105750	3673658625	2/3
22936954625	91747818500	1/4
56517825	75357100	3/4
19592846	97964230	1/5
3327950899994	8319877249985	2/5
3437223234	5728705390	3/5
104677490484	130846863105	4/5
19150763710393	114904582262358	1/6
2745064044632305	3294076853558766	5/6
3975350	27827450	1/7
936110884878	3276388097073	2/7
26869428369750	62695332862750	3/7
4813895358057500	8424316876600625	4/7
329402537360	461163552304	5/7
54709453541096250	63827695797945625	6/7
3305810795625	26446486365000	1/8
113394176313	302384470168	3/8
689223517385	1102757627816	5/8
978549117961625	1118341849099000	7/8
274817266734250	2473355400608250	1/9
41793444127641250	188070498574385625	2/9
176590156053048868	397327851119359953	4/9
6143093188763230	11057567739773814	5/9
601306443010000	773108283870000	7/9
6758920534667005000	7603785601500380625	8/9
104372894488263401	1043728944882634010	1/10
458710390065569889	1529034633551899630	3/10
8357399286061919849	11939141837231314070	7/10
49927726291701142521	55475251435223491690	9/10
11221334146768	123434675614448	1/11
4801442438	26407933409	2/11
33528490382546250	122937798069336250	3/11
5247317639775500	14430123509382625	4/11
1712007269488880	3766415992875536	5/11
13496488877215427538	24743562941561617153	6/11
587831133723750	923734638708750	7/11
58661465201996135000	80659514652744685625	8/11
2046772976463486000	2501611415677594000	9/11
414446414697850990	455891056167636089	10/11

Table 15. Table 15. Bhaskara pairs with a > 0 𝑎 0 a>0 , b ≤ 3 × 10 10 𝑏 3 superscript 10 10 b\leq 3\times 10^{10} after scanning the u / k 𝑢 𝑘 u/k ratios up to denominators k ≤ 200 ′ ​ 000 𝑘 superscript 200 ′ 000 k\leq 200^{\prime}000 . [ 14 , A106320]

$a$	$b$	$u/k$	s
2	2	1	1
128	128	1	2
625	1250	1/2	1
1458	1458	1	3
8192	8192	1	4
31250	31250	1	5
40000	80000	1/2	2
93312	93312	1	6
235298	235298	1	7
524288	524288	1	8
455625	911250	1/2	3
1062882	1062882	1	9
2000000	2000000	1	10
3543122	3543122	1	11
2560000	5120000	1/2	4
5971968	5971968	1	12
9653618	9653618	1	13
3430000	10290000	1/3	1
15059072	15059072	1	14
9765625	19531250	1/2	5
22781250	22781250	1	15
3975350	27827450	1/7	1
33554432	33554432	1	16
48275138	48275138	1	17
28130104	52743945	8/15	1
29160000	58320000	1/2	6
68024448	68024448	1	18
56517825	75357100	3/4	1
94091762	94091762	1	19
19592846	97964230	1/5	1
128000000	128000000	1	20
73530625	147061250	1/2	7
171532242	171532242	1	21
226759808	226759808	1	22
296071778	296071778	1	23
163840000	327680000	1/2	8
382205952	382205952	1	24
488281250	488281250	1	25
617831552	617831552	1	26
219520000	658560000	1/3	2
332150625	664301250	1/2	9
774840978	774840978	1	27
963780608	963780608	1	28
1189646642	1189646642	1	29
625000000	1250000000	1/2	10
1458000000	1458000000	1	30
1775007362	1775007362	1	31
254422400	1780956800	1/7	2
2147483648	2147483648	1	32
1107225625	2214451250	1/2	11
920414222	2235291682	7/17	1
2582935938	2582935938	1	33
3089608832	3089608832	1	34
1800326656	3375612480	8/15	2
2449105750	3673658625	2/3	1
3676531250	3676531250	1	35
1866240000	3732480000	1/2	12
4353564672	4353564672	1	36
3617140800	4822854400	3/4	2
5131452818	5131452818	1	37
3437223234	5728705390	3/5	1
6021872768	6021872768	1	38
3016755625	6033511250	1/2	13
1253942144	6269710720	1/5	2
7037487522	7037487522	1	39
2500470000	7501410000	1/3	3
8192000000	8192000000	1	40
4705960000	9411920000	1/2	14
9500208482	9500208482	1	41
10978063488	10978063488	1	42
9725113750	11493316250	11/13	1
12642726098	12642726098	1	43
7119140625	14238281250	1/2	15
14512627712	14512627712	1	44
16607531250	16607531250	1	45
18948593792	18948593792	1	46
2898030150	20286211050	1/7	3
10485760000	20971520000	1/2	16
21558430658	21558430658	1	47
24461180928	24461180928	1	48
4801442438	26407933409	2/11	1
27682574402	27682574402	1	49

Equations

$$a^2+b^2=x^3\wedge a^3+b^3=y^2$$

$$b^2=x^3\wedge b^3=y^2.$$

$$2a^2=x^3\wedge 2a^3=y^2.$$

$$a^2=4{\alpha }^3\wedge a^3=2{\beta }^2.$$

$$(1+u^2/k^2)b^2=x^3\wedge (1+u^3/k^3)b^3=y^2;$$

$$(u^2+k^2)b^2=k^2{x}^3\wedge (u^3+k^3)b^3=k^3{y}^2.$$

$$u^2+k^2$$

$$u^3+k^3$$

$$b$$

$$k$$

$$x$$

$$y$$

$$c_1\in \{0,1\}.$$

$$p_i\equiv 1(\mathrm{mod}4),\mathrm{if}{c}_i>0\wedge p_i\ge 3.$$

$$c_i+2b_i$$

$$d_i+3b_i$$

$$2b_i$$

$$3b_i$$

$$3c_i+6b_i=6k_i+9x_i\wedge 2d_i+6b_i=6k_i+4y_i$$

$$6b_i$$

$$(u^2+k^2)\frac{b^2}{k^2}=x^3\wedge (u^3+k^3)\frac{b^3}{k^3}=y^2$$

$$\bar{b}\equiv b/k.$$

$$u^2\equiv -k^2(\mathrm{mod}{p}_i),\mathrm{or}\mathrm{even}(\mathrm{mod}{p}_i^2),$$

$$u^3\equiv -k^3(\mathrm{mod}{p}_i).$$

$$u^2+k^2=x^3\wedge u^3+k^3=y^2,(u,k)=1$$

$$u^2\equiv -k^2(\mathrm{mod}{p}_i^3).$$

$$u^2+k^2=x^3,(u,k,x)=1,u,k,x\in \mathbb{Z}$$

$$\{u,k,x\}=\{s(s^2-3t^2),t(3s^2-t^2),t^2+s^2\}$$

$$(1+k,1+k^{2})=\left\{\begin{array}[]{ll}2,&2\nmid k;\\ 1,&2\mid k.\end{array}\right.$$

$$\frac{k^2+1}{k+1}=k-1+\frac{2}{k+1}$$

$$(u^{2}+k^{2},u^{3}+k^{3})=\left\{\begin{array}[]{ll}2,&2\mid(k-u);\\ 1,&2\nmid(k-u).\end{array}\right.$$

$$\frac{k^3+u^3}{k^2+u^2}=k+\frac{u^2(k-u)}{k^2+u^2},$$

$$\frac{k^2+u^2}{k}=k+\frac{u^2}{k},$$

Full text

1. Pair of Coupled Nonlinear Diophantine Equations

1.1. Scope

Following a nomenclature of Gupta we define [4, §4.4]:

Definition 1**.**

(Bhaskara pair) A Bhaskara pair is a pair $\{a,b\}$ of integers that solve the system of two nonlinear Diophantine equations of Fermat type:

[TABLE]

for some pair $\{x,y\}$.

Remark 1**.**

Lists of $a$ and $b$ are gathered in the Online Encyclopedia of Integer Sequences [14, A106319,A106320].

The symmetry swapping $a$ and $b$ in the equations indicates that without loss of information we can assume $0\leq a\leq b$, denoting the larger member of the pair by $b$.

We will not look into solutions where $a$ or $b$ are rational integers (fractional Bhaskara pairs).

The two equations can be solved individually [1, 5, 2].

Algorithm 1**.**

Given any solution $\{a,b\}$, further solutions $\{as^{6},bs^{6}\}$ are derived by multiplying both $a$ and $b$ by a sixth power of a common integer $s$, multiplying at the same time on the right hand sides $x$ by $s^{4}$ and $y$ by $s^{9}$.

Definition 2**.**

(Fundamental Bhaskara Pair) A fundamental Bhaskara pair is a Bhaskara pair $\{a,b\}$ where $a$ and $b$ have no common divisor which is 6-full—meaning there is no prime $p$ such that $p^{6}\mid a$ and $p^{6}\mid b$.

Although fundamental solutions are pairs that do not have a common divisor that is a non-trivial sixth power, individually $a$ or $b$ of a fundamental pair may contain sixth or higher (prime) powers.

Example 1**.**

The following is a fundamental Bhaskara pair with $2^{6}\mid a$, $2^{6}\nmid b$: $a=2^{6}\times 5^{4}\times 31^{3}\times 61^{3}$, $b=5^{4}\times 31^{3}\times 61^{3}\times 83$, $x=5^{3}\times 13\times 31^{2}\times 61^{2}$, and $y=3\times 5^{6}\times 7\times 31^{5}\times 61^{5}$.

2. Trivial Solutions

2.1. Primitive Solutions

A first family of solutions is found by setting $a=0$. This reduces the equations to

[TABLE]

$x^{3}$ must be a perfect cube, so in the canonical prime power factorization of $x^{3}$ all exponents of the primes must be multiples of three. Also in the canonical prime power factorization of $b^{2}$ all exponents must be even. So the first equation demands that the exponents on both sides must be multiples of $[2,3]=6$.

Definition 3**.**

Square brackets $[.,.]$ denote the least common multiple. Parenthesis $(.,.)$ denote the greatest common divisor.

In consequence all $b$ must be perfect cubes. Likewise the second equation demands that the exponents of $b^{3}$ and of $y^{2}$ are multiples of 6. In consequence all $b$ must be perfect squares. Uniting both requirements, all $b$ must be perfect sixth powers. And this requirement is obviously also sufficient: perfect sixth powers [14, A001014] generate Bhaskara pairs:

Theorem 1**.**

All integer pairs $\{0,n^{6}\}$, $n\in{\mathbb{Z}}_{0}$, are Bhaskara pairs. The associated right hand sides are $x=n^{4}$, $y=n^{9}$.

2.2. Bhaskara Twins

Definition 4**.**

(Bhaskara Twins) Bhaskara twins are a Bhaskara pair where $a=b$.

According to Definition 1 the Bhaskara twins [14, A106318] solve

[TABLE]

Working modulo 2 in the two equations requires that $x^{3}$ and $y^{2}$ are even, so $x$ and $y$ must be even, say $x=2\alpha$, $y=2\beta$. So

[TABLE]

The first equation requires by the right hand side that in the canonical prime power factorization of both sides the exponents of the odd primes are multiples of 3 and that the exponent of the prime 2 is $\equiv 2\pmod{3}$. By the left hand side of the first equation it requires that all exponents are even. So the exponents of the odd primes are multiples of 6, and the exponent of 2 is $\equiv 2\pmod{6}$. So from the first equation $a=2^{1+3\times}3^{3\times}5^{3\times}\cdots$, which means $a$ is twice a third power.

Definition 5**.**

The notation $3\times$ in the exponents means “any multiple of 3.”

The second equation in (4) demands by the right hand side that the exponents of the odd primes are even and that the exponent of 2 is $\equiv 1\pmod{2}$. Furthermore by the left hand side all exponents are multiples of 3. This means all exponents of the odd primes are multiples of 6, and the exponent of the prime 2 is $\equiv 3\pmod{6}$ So from the second equation $a=2^{1+2\times}3^{2\times}5^{2\times}\cdots$, which means $a$ must be twice a perfect square. Uniting both requirements, $a$ must be twice a sixth power. Obviously that requirement is also sufficient to generate solutions:

Theorem 2**.**

The Bhaskara Twins are the integer pairs $\{2n^{6},2n^{6}\}$, $n\in{\mathbb{Z}}_{0}$. The associated free variables are $x=2n^{4}$, $y=4n^{6}$.

3. Rational Ratios of the two Members

3.1. Prime Factorization

The general solution to (1) is characterized by some ratio $a/b=u/k\leq 1$ with some coprime pair of integers $(k,u)=1$. Cases where $u$ and $k$ are not coprime are not dealt with because they do not generate new solutions.

If $k$ were not a divisor of $b$, $a=ub/k$ would require that $k$ is a divisor of $u$ to let $a$ be integer, contradicting the requirement that $u$ and $k$ are coprime.

Algorithm 2**.**

We only admit the denominators $k\mid b$.

Theorem 1 and 2 cover the solutions of the special cases $u=0$ or $u=1$. Introducing the notation into (1) yields

[TABLE]

[TABLE]

Define prime power exponents $c_{i}$, $d_{i}$, $b_{i}$, $x_{i}$ and $y_{i}$ as follows by prime power factorizations, where $p_{i}$ is the $i$-th prime:

[TABLE]

In (7), $u^{2}+k^{2}$ is the sum of two squares [14, A000404]. Because $u$ and $k$ are coprime, these $u^{2}+k^{2}$ are 2, 5, 10, 13, 17, 25, 26, 29, 34, 37, 41,…, numbers whose prime divisors are all $p\equiv 1\pmod{4}$ with the exception of a single factor of $2$ [14, A008784][12, Thm. 2.5][9, Thm. 3]:

Lemma 1**.**

[TABLE]

[TABLE]

Example 2**.**

If $u=2^{6}$, $k=83$ as in Example 1, $u^{2}+k^{2}=5\times 13^{3}$, so $c_{3}=1$, $c_{6}=3$, and $u^{3}+k^{3}=3^{2}\times 7^{2}\times 31\times 61$, so $d_{2}=2$, $d_{4}=2$, $d_{11}=1$, $d_{18}=1$.

The uniqueness of the prime power representations in (6) requires for all $i\geq 1$

[TABLE]

for unknown sets of ${b_{i},x_{i},y_{i}}$ and known ${c_{i},d_{i},k_{i}}$ (if $u/k$ is fixed and known). For some $i$—including all $i$ larger than the index of the largest prime factor of $[u^{2}+k^{2},u^{3}+k^{3},k]$ once $u/k$ is fixed—we have $c_{i}=d_{i}=k_{i}=0$. For these

[TABLE]

The first equation requires $2\mid x_{i}$ and $3\mid b_{i}$. The second equation requires $3\mid y_{i}$ and $2\mid b_{i}$. The combination requires $6\mid b_{i}$. The absence of the $i$-th prime allows to multiply $b$ by a sixth (or 12th or 18th…) power of the $i$-th prime. These factors are of no interest to the construction of fundamental Bhaskara pairs.

In practice we use the Chinese Remainder Theorem (CRT) for all $i$, whether the $c_{i}$ or $d_{i}$ are zero or not [13, 7]. Multiply (15a) by 3 and (15b) by 2,

[TABLE]

such that the two factors in front of the $b_{i}$ are the same, and work modulo 9 in the first equation and modulo 4 in the second:

[TABLE]

Because 9 and 4 are relatively prime, the CRT guarantees that an integer $6a_{i}$ exists. Furthermore the result will always be a multiple of 6 (hence $a_{i}$ an integer), because from (18a) the equations read modulo 3 we deduce that $6a_{i}$ is a multiple of 3, and from (18b) read modulo 2 that $6a_{i}$ is a multiple of 2:

Algorithm 3**.**

For each ratio $a/b=u/k$, the prime power decompositions of $u^{2}+k^{2}$ and $u^{3}+k^{3}$ generate a unique exponent $b_{i}$ of the prime power $p_{i}^{b_{i}}$ of a conjectured solution $b$.

We compute $6b_{i}\pmod{9\times 4}$ by any algorithm [11], so $b_{i}$ is determined $\pmod{6}$.

The values of $b_{i}-k_{i}$ that result from the CRT for the three relevant values of $c_{i}$ and the two relevant $d_{i}$ establish Table 2. The rows and columns are bi-periodic for both $c_{i}$ and $d_{i}$; the entries depend only on $d_{i}\pmod{2}$ and on $c_{i}\pmod{3}$. The zero at the top left entry where $d_{i}$ is a multiple of 2 and $c_{i}$ a multiple of 3 means that a prime $p_{i}$ is “discarded” and its associated sixth power shoved into the $x^{3}$ and $y^{2}$ in equation (6). That zero in the table purges the non-fundamental solutions.

Algorithm 4**.**

For any fraction $u/k$ of the Farey tree with $(u,k)=1$, construct the set $\{p_{i}\}$ of common prime factors of $k$, $u^{2}+k^{2}$ and $u^{3}+k^{3}$. Compute the exponents $k_{i}$, $c_{i}$ and $d_{i}$ of their prime power factorizations. Construct for each $i$ the exponent $b_{i}$ as the sum of the entry in Table 2 plus $k_{i}$, and compose $b=\prod_{i}p_{i}^{b_{i}}$.

Remark 2**.**

$u^{2}+k^{2}$* and $u^{3}+k^{3}$ have no common divisor larger than 2 (see Lemma 5 in the Appendix). So the only case where $c_{i}$ and $d_{i}$ are both nonzero may occur at prime index $i=1$ and if $u$ and $k$ are both odd. For that reason Table 2 never fathers odd prime powers $p^{1}$ or $p^{5}$, and the only odd prime powers in $b$ of that form are those contributed by the factor $k=\prod_{i}p_{i}^{k_{i}}$.*

Lemma 2**.**

Because $k$ has no common prime factors with either $u^{2}+k^{2}$ or $u^{3}+k^{3}$ according to Lemma 6 in the Appendix, nonzero $k_{i}$ appear only where $c_{i}=d_{i}=0$.

This ensures that in the construction of $b$ all $p_{i}^{k_{i}}$ appear as factors and that $k\mid b$. $a=ub/k$ generated by the algorithm is always an integer.

The step from (15)—necessary and sufficient for a solution—to (18) eliminates $x_{i}$ and $y_{i}$ by applying a modular sieve; the modular sieve reduces (18) to a necessary condition. To show that these $b$ are also sufficient and indeed solve the coupled Diophantine equations, the step from (15) to (18) must be reversible, such that all solutions of (18) also fulfill (15). Indeed we can find a multiple of 9 and add it to the right hand side of the equivalence (18a) such that it becomes an equality, and we can find a multiple of 4 and add it to the right hand side of the equivalence (18b) such that it becomes an equality. Dividing the two equations by 3 and 2, respectively, turns out to be a constructive proof that the $3x_{i}$ and $2y_{i}$ exist, and that they are multiples of $3$ and $2$:

Theorem 3**.**

For each given ratio $a/b=u/k$, the Algorithm 4 generates a unique fundamental solution $b$.

Lemma 2 means that the data reduction of (6) effectively deals only with

[TABLE]

with three integers $u^{2}+k^{2}$, $u^{3}+k^{3}$ and

[TABLE]

Can we generate more solutions by not just copying the prime factors of $k$ over to $b$ but introducing higher exponents, such that $b_{i}-k_{i}>0$? The prime power decomposition of (19) would demand that the surplus factor $p_{i}^{2(b_{i}-k_{i})}$ divides $x^{3}$ and that the surplus factor $p_{i}^{3(b_{i}-k_{i})}$ divides $y^{2}$. Lemma 2 ensures that these are the only contributions to $x_{i}^{3}$ and $y_{i}^{2}$, so effectively $b_{i}-k_{i}$ must be multiples of 6. These sixth powers are introduced at the same time to $a=ub/k$; so that deliberation does not generate any other fundamental pairs. With a similar reasoning, multiplying $b$ by any prime power of a prime that is not a prime factor of $k$—but coupled to $c_{i}\pmod{3}$ and to $d_{i}\pmod{2}$ via (15)—admits only further exponents that are multiples of 6, and again there is no venue for any other fundamental solutions from that subset of prime factors. The solutions are indeed unique as claimed by Theorem 3.

3.2. Examples with $u=1$

The algorithm and results will be illustrated for a set of small $1/k$ and integer ratios $b/a$ in Tables 3–8. The tables have 4 columns, the prime index $i$, the exponents $c_{i}$, $d_{i}$ and $k_{i}$ defined by the prime factorization of $u^{2}+k^{2}$, of $u^{3}+k^{3}$, and of $k_{i}$, and the factor $p_{i}^{b_{i}}$ generated by the CRT. “Spectator” primes, the cases (rows) where $c_{i}=d_{i}=k_{i}=0$, are not tabulated; they would be absorbed in the sixth powers of non-fundamental solutions.

3.2.1. u/k=1

The case $u=k=1$ in Table 3 reconvenes the Bashkara Twin Pairs of Theorem 2.

3.2.2. u/k=1/2

Looking at the second line of Table 1 we have only contributions for primes $p_{2}=3$ and $p_{3}=5$ in Table 4.

From there all solutions of the form $\{a=b/2,b\}$ are given by the set of $b=2\times 5^{4}s^{6}$ with non-negative integers $s$, where $\{x,y\}=\{5^{3}s^{4},3\times 5^{6}s^{9}\}$.

3.2.3. $u/k=1/3$

From the line $k=3$ of Table 1 we have the contribution from the prime factors of Table 5.

3.2.4. $k\geq 4$

The primes of the line $k=4$ of Table 1 generate Table 6.

Further solutions $(a=b/k,b)$ with $u/k=1/5\ldots 1/6$ are gathered in Tables 7–8.

3.3. Examples with $u>1$

Some cases where the numerator of $u/k$ is $u>1$ and therefore $b$ not an integer multiple of $a$ are illustrated in Tables 9–13.

4. Table of Fundamental Solutions

Systematic exploration of ratios $u/k$ sorted along increasing $k$ generates Table 14.

The rather larger value of $b$ for $u/k=5/6$ is derived with Table 10 from the fact that $u^{2}+k^{2}$ have a rather large isolated prime factor ($p_{18}=61$) which enters with its fourth power.

The rather small value of $b$ at $u/k=2/11$ is explained with Table 11 from the fact that $u^{2}+k^{2}$ is a cube, which does not contribute to $b$ at all because the exponent is zero for $c_{i}\equiv 0\pmod{3}$, $d_{i}\equiv 0\pmod{2}$ in Table 2.

Multiplications of solutions of Table 14 with common powers $s^{6}$ and sorting along increasing $b$ leads to Table LABEL:tab.res. Trivial solutions with $a=0$ ($u/k=0$) are not listed. The fundamental solutions are flagged by $s=1$ and indicate where Table 14 intersects with Table LABEL:tab.res.

Remark 3**.**

The list in Table LABEL:tab.res is not proven to be complete up to its maximum $b$, because only a limited number of ratios $a/b=u/k$ were computed.

5. Criteria On The Larger Member

5.1. Brute Force

Building a complete table of the $b$ that are solutions up to some maximum calls for an efficient method to decide whether any candidate $b$ has an associate $a$ that solves the equations.

The brute force method is rather slow: one could check all individual $0\leq a\leq b$ whether the sum $a^{2}+b^{2}$ is a cube and whether $a^{3}+b^{3}$ is a square; this effort grows $\sim b$. A faster brute force method considers all cubes $x^{3}$ in the range $b^{2/3}$ up to $(2b)^{2/3}$, derives the associates $a=\sqrt{x^{3}-b^{2}}$ and checks these first whether they are integer and then whether they solve the equations; this effort grows $\sim b^{2/3}$.

5.2. Removal of Non-fundamental Pairs

Reverse engineering the results of the previous sections starts from the the prime power decomposition of $b$. The set of its factors $p_{i}^{b_{i}}$ has $\omega(b)$ members, where $\omega(.)$ denotes the number if distinct primes that divide the argument [14, A001221]. For any subset of the $p_{i}$ where $b_{i}\geq 6$, we can split off a set of sixth prime powers that define a factor $s^{6}$ considered a part of a non-fundamental solution, and continue to figure out whether $b/s^{6}$ is a member of a fundamental pair. For the rest of the section we only deal with this checking of $b$ as a member of a fundamental pair. Note that still the prime factor decomposition of $b$ may have prime exponents that are $\geq 6$.

5.3. Congruences for Fundamental Pairs

This set of prime powers of $b$ is divided in an outer decision loop in $2^{\omega(b)}$ different ways into two disjoint subsets; one subset defines the prime powers of $k=\prod_{i}p_{i}^{k_{i}}$, the other the prime powers of the conjugate $\bar{b}=p/k$, $\omega(\bar{b})=\omega(b)-\omega(k)$.

If the subset of the prime powers of $k$ is chosen to be empty, $k=u=1$, this reduces to a trivial check whether $b$ is a member of a Bhaskara Twin Pair of the format of Theorem 2.

For each of these candidates $k$ of $b$ we wish to decide whether an associate coprime $u$ exists that solves (19).

• •

If the prime power set of $\bar{b}$ contains exponents $\equiv\pm 1\pmod{6}$, we reject the $k$, because (see Remark 2) it is impossible to find coprime $u^{2}+k^{2}$ and $u^{3}+k^{3}$ that complement them to cubes and squares. (To reject means to book them as not fostering solutions.)

• •

If the prime power set of $\bar{b}$ contains exponents $\geq 6$ we reject the $k$ because the same prime power appears in $a=u\bar{b}$ which violates the search criterion for fundamental pairs.

5.3.1.

The prime power set of $\bar{b}$ now contains primes with exponent 2, 3 or 4. According to Table 2 the exponent 2 enforces that the prime factor $p_{i}^{1+3\times}$ appears in $u^{2}+k^{2}=\prod_{i}p_{i}^{c_{i}}$ to complement $x^{3}$, the exponent 4 enforces that the prime factor $p_{i}^{2+3\times}$ appears in $u^{2}+k^{2}$ to complement $x^{3}$, and the exponent 3 enforces that the prime factor $p_{i}^{1+2\times}$ appears in $u^{3}+k^{3}=\prod_{i}p_{i}^{d_{i}}$ to complement $y^{2}$.

• •

We reject exponent sets $\{c_{i}\}$ if they violate Lemma 1.

This knowledge that some specific primes or prime powers appear in the prime power factorization of $u^{2}+k^{2}$ or $u^{3}+k^{3}$ is used to narrow down the search set of $u$ because for these known $p_{i}$ and given $k$ the quadratic and cubic residues must be

[TABLE]

respectively

[TABLE]

5.3.2.

The worst case of the analysis occurs if the entire set of prime powers of $b$ is packed into $k$, $k=b$. Then $\bar{b}=1$ and none of the rejection criteria above applies. We are facing the original set of equations just with the additional support information that $k$ is known and that $u$ and $k$ need to be coprime:

[TABLE]

Remark 4**.**

The solutions $k$ for the first equation are [14, A282095]; the solutions $k$ for the second equation are [14, A282639]. The task is to find the values that are in both sequences.

It is unknown whether any solutions to (23)—coprime Bhaskara pairs—exist.

According to Remark 5 the parities of $k$ and $u$ differ, so $u^{2}+k^{2}$ is odd. In any case the prime factors of $x$ are restricted by Lemma 1 and appear with exponents that are multiples of 3; the prime factor 2 does not appear. The prime factors of $k$ are known, and the prime factor set of $u$ is restricted by not intersecting the prime factor set of $k$. A weak upper limit of the largest prime factor in $u$ is $k$; a weak upper limit of the largest prime factor in $x$ is $(2k^{2})^{1/3}$. $u$ and $x$ have no common prime factor (because that would need to appear also in $k$ and violate co-primality). Similarly $k$ and $x$ have no common prime factor.

The simplest way to implement a sieve is to work in a loop over hypothetical prime factors $p_{i}|x$ and discard them if $-k^{2}$ are not quadratic residues as required by (23):

[TABLE]

A support for brute force construction of all solutions to the first equation in (23)—faster than a loop over all coprime $u$—is given by:

Lemma 3**.**

[3, 5]** A solution to

[TABLE]

satisfies

[TABLE]

for some $s,t\in\mathbb{Z}$ with $(s,t)=1$ and $st\neq 0$.

Algorithm 5**.**

Loop over all divisors $t$ (of both signs) of $k$, compute the conjugate divisor $k/t=3s^{2}-t^{2}$. Check that $s$ is integer, else discard $t$. If $s$ is not coprime to $t$, discard $t$. Compute $u=s(s^{2}-3t^{2})$ and take the absolute value. If that absolute value is larger than $k$ or not coprime to $k$, discard $t$, otherwise a solution of (25) is found.

Remark 5**.**

The parities of $s$ and $t$ in (26) are different. In detail: If $k$ is

• •

odd, all divisors $t$ are odd, and the conjugate $3s^{2}-t^{2}$ are also odd. So $3s^{2}$ are even. Therefore $s^{2}$ must be even and eventually $s$ be even. The conjugate $s^{2}-3t^{2}$ are odd and $u$ are even.

• •

even, and $t$ is even: Because we request $s$ to be coprime to $t$, $s$ must be odd, so $3s^{2}$ is odd, and the conjugate $3s^{2}-t^{2}$ is odd. The conjugate $s^{2}-3t^{2}$ is odd, and $u$ is odd.

• •

even and $t$ is odd, the conjugate $3s^{2}-t^{2}$ must be even, so $3s^{2}$ must be odd and hence $s$ must be odd. Its conjugate $s^{2}-3t^{2}$ is even, so $u$ is even. This violates $(u,k)=1$ and does not occur.

6. Summary

We have shown that for each ratio $a/b$ a unique smallest (fundamental) solution of the non-linear coupled diophantine equations (1) exists, which can be constructed by modular analysis via the Chinese Remainder Theorem. We constructed these explicitly for a set of small ratios.

Appendix A Greatest Common Divisors

Lemma 4**.**

The greatest common divisor of $1+k$ and $1+k^{2}$ is

[TABLE]

Proof.

The Euclidean Algorithm to construct the greatest common divisor starts with [8]

[TABLE]

and basically terminates at this step, so $(1+k,1+k^{2})=(1+k,2)$. This is obviously $1$ or $2$ for even and odd $k$ as claimed. ∎

Lemma 5**.**

The greatest common divisor of $u^{2}+k^{2}$ and $u^{3}+k^{3}$ for coprime $(u,k)=1$ is

[TABLE]

Proof.

The first step of the Euclidean Algorithm is

[TABLE]

so $(u^{3}+k^{3},u^{2}+k^{2})=(u^{2}(k-u),k^{2}+u^{2})$. Assume $p^{j}$ is one of the inquired common prime power factors of the common divisor such that $p^{j}\mid(u^{2}(k-u))$ and $p^{j}\mid(k^{2}+u^{2})$, say $k^{2}+u^{2}=vp^{j}$ for some $j>0$, $v>0$. The first requirement induces $p\mid u^{2}$ or $p|(k-u)$.

• •

Suppose $p\mid u^{2}$, then $p\mid u$ by the uniqueness of prime factorizations, say $u=\alpha p$. Insertion of this into $k^{2}+u^{2}=vp^{j}$ and evaluating both sides modulo $p$ leads to the requirement $k^{2}\equiv 0\pmod{p}$, therefore $p\mid k$. This contradicts the requirement $p\mid u$ because $k$ and $u$ are coprime and must not have a common factor $p$. In conclusion $p\nmid u^{2}$.

• •

Since $p\nmid u^{2}$, $p^{j}|(u^{2}(k-u))$ requires $p^{j}\mid(k-u)$. Rewrite $k^{2}+u^{2}=2u^{2}-2u(k-u)+(k-u)^{2}=vp^{j}$. Working modulo $p^{j}$ this becomes $2u^{2}\equiv 0\pmod{p^{j}}$. Since $p$ does not divide $u^{2}$ as shown in the previous bullet, this requirement reduces to $2\equiv 0\pmod{p^{j}}$, leaving $p^{j}=2^{1}$ as the only common prime divisor candidate.

It is furthermore obvious that for odd $k$ and odd $u$ both $u^{2}+k^{2}$ and $u^{3}+k^{3}$ are even, so the common prime factor $2$ is indeed achieved. ∎

Lemma 6**.**

If $(u,k)=1$, $k$ is coprime to $u^{2}+k^{2}$ and coprime to $u^{2}+k^{3}$.

Proof.

In the first case the first step of the Euclidean Algorithm to compute $(k^{2}+u^{2},k)$ is

[TABLE]

in the second case

[TABLE]

So the greatest common divisors are $(k^{2}+u^{2},k)=(u^{2},k)$ and $(k^{3}+u^{3},k)=(u^{3},k)$. Both expressions equal 1 because we assume that $u$ and $k$ are coprime. ∎

Appendix B Sum of two Squares

Lemma 7**.**

There are no solutions to $4k+1+p^{2}=x^{3}$ with $p$ a prime and $4k+1<p^{2}$.

Proof.

This is obvious for the even prime where $k=0$ is the only candidate. The other primes are either of the form $p=4m+1$ with $4k+1+p^{2}=2(1+2k+4m+8m^{2})$ or $q=4m+3$ with $4k+1+q^{2}=2(5+2k+12m+8m^{2})$. In any case $4k+1+p^{2}$ is two times an odd number for odd primes $p$. Because $(4k+1,p)=1$, Lemma 1 applies and the 2 must appear on the right hand side either not at all or risen to the first power. Both contradicts the request for a perfect cube $x^{3}$ on the right hand side. ∎

Lemma 8**.**

There are no solutions to

[TABLE]

where $p$ is a prime, $(u,p)=1$ and $1\leq u\leq p$.

Proof.

The case of the even prime is obvious because $1+2^{2}$ is not a cube, and the case of the only prime with $3\mid p$ is also obvious because $1+3^{2}$ and $2+3^{2}$ are not cubes. The proof is based on the failure to create any of the parameterizations required by Lemma 3 considering all $t|p$ one by one:

• •

$t=1$ leads to the conjugate divisors $p^{2}/t=p^{2}=3s^{2}-1$. The other primes fall into the categories $p=3m+1$ where $p^{2}\equiv 1\pmod{3}$ and $q=3m+2$ where $q^{2}=1\pmod{3}$. This contradicts $p^{2}\equiv-1\pmod{3}$ of the conjugate required above, so there are no solutions induced by $t=1$.

• •

$t=-1$ leads to a conjugate $p^{2}/t=-p^{2}$ which is negative and cannot be equal to the (essentially) positive $3s^{2}-1$.

• •

$t=p$ leads to the conjugate divisor $p^{2}/p=p=3s^{2}-p^{2}$, $p+p^{2}=p(p+1)=3s^{2}$. For primes of the form $p=3m+1$ we have $p(p+1)\equiv 2\pmod{3}$ and for primes of the form $q=3m+2$ we have $q(q+1)\equiv 0\pmod{3}$. So only the primes $\equiv 2\pmod{3}$ generate $3s^{2}$ that are multiples of 3. If $q=3m+2$ then $q(q+1)=3(m+1)(3m+2)$, so we require $s^{2}=(m+1)(3m+2)$. Because $m+1$ and $3m+2$ are coprime, their product can only be a perfect square $s^{2}$ if $m+1$ and $3m+2$ are individually perfect squares, say $m+1=\alpha^{2}$, $3m+2=\beta^{2}$, $(\alpha,\beta)=1$. $\beta^{2}-3\alpha^{2}=-1$. This negative Pell equation with $D=3$ is not solvable [10, 6]; the parameterization does not generate solutions.

• •

$t=-p$ leads to the conjugate $p^{2}/p=-p=3s^{2}-p^{2}$. $p(p-1)=3s^{2}$ with $p=3m+1$ implies $p(p-1)\equiv 0\pmod{3}$. $q(q-1)=3s^{2}$ with $q=3m+2$ implies $q(q-1)\equiv 2\pmod{3}$. So only primes $p=3m+1$ remain candidates to represent $3s^{2}$, and then $p(p-1)=3m(3m+1)=3s^{2}$ requires $s^{2}=m(3m+1)=mp$. Because $m$ and $3m+1$ are coprime, this requires that $p\mid s$. Because $s$ is a divisor of $u$, this violates the requirement that $(u,p)=1$ and does not foster solutions.

∎