Solutions of the Exponential Equation 7x^2 + 59y^2 = 3^m. A simple Algorithm producing all the primitive solutions

TL;DR

This paper presents an elementary arithmetic-based method and a simple algorithm to find all primitive positive solutions of the exponential Diophantine equation 7x^2 + 59y^2 = 3^m, which can be generalized to similar equations.

Contribution

It introduces a novel elementary approach and an efficient algorithm for solving a specific exponential Diophantine equation, avoiding radicals or complex numbers.

Findings

Algorithm generates all primitive solutions

Method applies to a class of similar equations

No radicals or complex numbers used

Abstract

We provide a method, using essentially elementary arithmetic, to solve the exponential diophantine equation 7x^2 + 59y^2 = 3^m, which leads to a simple algorithm, with no use of radicals or complex numbers, that generates all the (infinitely many) primitive and positive solutions of this equation. The method can be generalized to solve a class of exponential equations of the form ax^2 + by^2 = ck^z.

Equations

$$7x^2+59y^2=3^m$$

$$m=10k+5,k=0,1,2,3,\dots$$

$$(1,2,5),(701,430,15),(262009,78842,25),(78606773,10718566,35).$$

$$A(p,q)$$

$$B(p,q)$$

$$C(p,q)$$

$$x=3^{-5}|A(p,q)|,y=3^{-5}|B(p,q)|,x^{\prime }=|A(-p,q)|,y^{\prime }=|B(-p,q)|.$$

$$5,15,25,35,45,55,65,75,\dots$$

$$7(59p^2-236pq-7q^2{)}^2+59(-118p^2-14pq+14q^2{)}^2=3[9(59p^2+7q^2){]}^2$$

$$7A^2+59B^2=3C^2$$

$$7A^2+59B^2=3^{2\omega +5}$$

$$\gcd (u,v)=3^r,r\ge 1$$

$$\gcd (u,v)=3^r,1\le r\le 5$$

$$u=3^r\theta ,\text{where}1\le r\le 4\text{and}3\nmid \theta$$

$$7u^2+59(3^5\lambda -2u{)}^2=3^m$$

$$7u^2+3^{10}\cdot 59{\lambda }^2+236u^2-3^5\cdot 236\lambda u=3^m$$

$$3^5{u}^2+3^{10}\cdot 59{\lambda }^2-3^5\cdot 236\lambda u=3^m$$

$$3^{5+2r}\cdot {\theta }^2+3^{10}\cdot 59{\lambda }^2-3^{5+r}\cdot 236\lambda \theta =3^m$$

$$3^r{\theta }^2+3^{5-r}\cdot 59{\lambda }^2-236\lambda \theta =3^{m-5-r}$$

$$2u+v$$

$$=-486pq=3^5(-2pq),\text{where}3\nmid -2pq$$

$$7u^2+59v^2=3^{2\omega +5}$$

$$7(3^{-5}u{)}^2+59(3^{-5}v{)}^2=3^{2\omega -5}$$

$$7x^2+59y^2=3^{2\omega -5}$$

$$7[A(-p,q){]}^2+59[B(-p,q){]}^2=3^{2\omega +5}$$

$$\left(\frac{2}{59}\right)$$

$$\left(\frac{3}{59}\right)$$

$$\left(\frac{7}{59}\right)$$

$$\delta =\gcd (A,B)=\gcd (A,C)=\gcd (B,C).$$

$$2^1,3^s(1\le s\le 5),7^1\text{or}59^1$$

$$X=\frac{A(p,q)}{C(p,q)}\text{and}Y=\frac{B(p,q)}{C(p,q)}$$

$$Y=\theta X+\frac{2-\theta }{9}$$

$$7(9X{)}^2+59(9\theta X+2-\theta {)}^2=243$$

Full text

1. (i)

The proof of the identity is straightforward. 2. (ii)

$7q^{2}+59p^{2}=3^{\omega}$   yields   $C=C(p,q)=9(7q^{2}+59p^{2})=3^{\omega+2}$.   Hence, $(i)$ yields

$7A^{2}+59B^{2}=3C^{2}=3(3^{\omega+2})^{2}=3^{2\omega+5}$.

∎

**Remark 2:

**Let $p,q\in\mathbb{Z}^{\ast}$. Then, $A(p,q)\neq 0$   and   $B(p,q)\neq 0$.

Indeed: set $u=A(p,q)$,  $v=B(p,q)$   and   $w=C(p,q)$.   Then, $w=9(59p^{2}+7q^{2})>0$,  and, by

part $(i)$ of lemma $1$, we have $7u^{2}+59v^{2}=3w^{2}$. Assuming $u=0$ would lead to the contradiction that $\sqrt{\frac{3}{59}}$ is rational, and assuming $v=0$ would lead to the contradiction that $\sqrt{\frac{3}{7}}$ is rational.

**Lemma 2:

**Let $u,v\in\mathbb{Z}^{\ast}$ and let $m$ be an integer, $m\geq 15$. Suppose that

1. (i)

$7u^{2}+59v^{2}=3^{m}$, where $3\mid u$. 2. (ii)

$2u+v=3^{5}\lambda$    for some $\lambda\in\mathbb{Z}^{\ast}$,   where $3\nmid\lambda$.

Then,   $\gcd(u,v)=3^{5}$.

[TABLE]

If we assume that $r\geq 6$, then $3^{6}$ would divide $2u+v$, and hence by $(ii)$, $3^{6}$ would divide $3^{5}\lambda$, so $3$ would divide $\lambda$, a contradiction. We conclude that

[TABLE]

Now, we claim that $r=5$: for the purpose of contradiction, suppose that $1\leq r\leq 4$. Under this assumption, $3^{5}$ cannot divide $u$: otherwise by $(i)$ and as $m\geq 15$, $3^{5}$ would divide $v$, hence $3^{5}$ would divide $\gcd(u,v)$.   We then set

[TABLE]

Using $(i)$ and $(ii)$ , we may write

[TABLE]

Expanding yields

[TABLE]

That is,

[TABLE]

Replacing $u$ by $3^{r}\theta$ (given by $(j)$) gives

[TABLE]

Dividing by $3^{5+r}$ yields

[TABLE]

Since $1\leq r\leq 4$   and   $m\geq 15$,   all the terms in $(\ell)$ are divisible by $3$, except the term $-236\lambda\theta$.

We obtain a contradiction. ∎

**Corollary 2:

**Let $(q,p,\omega)$ be a primitive solution of $(1)$, with $pq\equiv-1\pmod{3}$. Set $u=A(p,q)$ and $v=B(p,q)$. Then, $\gcd(u,v)=3^{5}$.

[TABLE]

By lemma $2$, we obtain that $\gcd(u,v)=3^{5}$. ∎

Proof of theorem 1:

Let $(q,p,\omega)$ be a pp-solution of $(1)$. As pointed out in theorem $1$, we may assume that $pq\equiv-1\pmod{3}$. Set $u=A(p,q)$ and $v=B(p,q)$. By corollary $2$, we have $\gcd(u,v)=3^{5}$. Set $x=3^{-5}|u|$ and $y=3^{-5}|v|$. Thus, $x$ and $y$ are coprime positive integers.

Since by hypothesis $7q^{2}+59p^{2}=3^{\omega}$, lemma $1$ $(ii)$ provides:

[TABLE]

Dividing by $3^{10}$ yields

[TABLE]

Hence,

[TABLE]

As $x>0$,   $y>0$   and   $\gcd(x,y)=1$,   we see that $(x,y,2w-5)$ is a pp-solution of $(1)$.

$\bullet$ Next, set $x^{\prime}=|A(-p,q)|$ and $y^{\prime}=|B(-p,q)|$.

With $pq\equiv-1\pmod{3}$, we may write

$A(-p,q)=59p^{2}+236pq-7q^{2}\equiv 59-236-7=-181\equiv-1\pmod{3}$.

Hence, $3\nmid x^{\prime}$. Now, lemma $1$ $(ii)$ applied to $-p$ and $q$ yields:

[TABLE]

Finally, we have $7x^{\prime 2}+59y^{\prime 2}=3^{2\omega+5}$, where $3\nmid x^{\prime}$.

Hence clearly, $(x^{\prime},y^{\prime},2\omega+5)$ is a pp-solution of $(1)$. $\square$

**Remark 3:

**In the light of what precedes and particularly of lemma $1$ $(ii)$, we end this section by providing a simple and useful criterion to "recognize" the successor, that we will use in section $6$.

$\bullet$ Given that $(x,y,m)$ and $(q,p,\omega)$ are known to be two primitive and positive solutions of $(1)$ (here, we need neither to suppose that $pq\equiv-1\pmod{3}$, nor to assume any relation between the exponents $m$ and $\omega$), then, the reader can check the following:

1. (i)

If we can just show that $x=|A(\pm p,q)|$ and $y=|B(\pm p,q)|$ (the "plus" is to be taken with the "plus", and the "minus" with the "minus"), then, $(x,y,m)$ is the second successor of $(q,p,\omega)$. 2. (ii)

If we can just show that $x=3^{-5}|A(\pm p,q)|$ and $y=3^{-5}|B(\pm p,q)|$ (the "plus" is to be taken with the "plus", and the "minus" with the "minus"), then, $(x,y,m)$ is the first successor of $(q,p,\omega)$.

5. Preliminaries for theorem 2:

Notation:

As usual, $\left(\frac{x}{y}\right)$ denotes Legendre’s symbol.

Recall Legendre’s theorem for ternary quadratic forms (see for ex. $[6]$):

**Legendre’s Theorem:

**Let $a,b,c$ be positive square-free integers, and pairwise coprime. Then, the equation $ax^{2}+by^{2}=cz^{2}$ has a non-trivial integer solution   if and only if   $\left(\frac{-ab}{c}\right)=\left(\frac{bc}{a}\right)=\left(\frac{ca}{b}\right)=+1$.

**Lemma 3:

**The equation $7x^{2}+59y^{2}=z^{2}$ has no non-trivial integer solution.

$\bullet$ Since $3^{m}$ has the form $z^{2}$ when $m$ is even, we obtain:

Corollary 3:

The equation $7x^{2}+59y^{2}=3^{m}$ has no integer solution with $m$ even.

Lemma 4:

The equation $7x^{2}+59y^{2}=2z^{2}$ has no non-trivial integer solution.

Lemma 5:

The equation $7x^{2}+59y^{2}=6z^{2}$ has no non-trivial integer solution.

We have $\left(\frac{6\times 7}{59}\right)=\left(\frac{2}{59}\right)\left(\frac{3}{59}\right)\left(\frac{7}{59}\right)$.

Now, using (Gauss’) law of quadratic reciprocity, etc. (see for ex. $[7]$ or $[8]$) we may write:

[TABLE]

Hence, $\left(\frac{6\times 7}{59}\right)=(-1)(+1)(+1)=-1$. ∎

$\bullet$ Since $2\cdot 3^{m}$ has the form $2z^{2}$ or $6z^{2}$ according to whether $m$ is even or odd, we obtain

(by lemmas $4$ and $5$) the following:

Corollary 4:

The equation $7x^{2}+59y^{2}=2\cdot 3^{m}$ ($m>0$) has no integer solution.

**Remark 4:

**Let $p,q\in\mathbb{Z}^{\ast}$. Set $A=A(p,q)$, $B=B(p,q)$ and $C=C(p,q)$.

Let $\delta$ denote the positive gcd of $A$ and $B$. According to lemma $1$ $(i)$, we have: $7A^{2}+59B^{2}=3C^{2}$. In virtue of this identity, it should be clear that:

[TABLE]

**Proposition 1:

**Let $p,q\in\mathbb{Z}^{\ast}$, $p$ and $q$ coprime. Set $A=A(p,q)$ and $B=B(p,q)$. Let $\delta$ denote the positive gcd of $A$ and $B$. Then,

1. (i)

$\delta$ divides $2\cdot 3^{5}\cdot 7\cdot 59$. 2. (ii)

We have:   $[\ 7\mid\delta\iff 7\mid p\ ]$   and   $[\ 59\mid\delta\iff 59\mid q\ ]$.

Proof:

1. (i)

If $\delta=1$, there is nothing to prove. We then assume that $\delta>1$. We show that every primary factor of $\delta$ must have one of the following forms:

[TABLE]

Indeed, let $\pi^{s}$, $s\geq 1$, be a primary factor of $\delta$. Then, $\pi^{s}$ divides $A$ and $B$. Hence, $\pi^{s}$ divides $2A+B=-2\cdot 3^{5}\cdot pq$. We consider two cases:

case 1: $\pi\nmid pq$:

Then, $\pi^{s}$ is coprime to $pq$. By Gauss’ theorem, $\pi^{s}$ divides $2\cdot 3^{5}$. Hence, either $\pi=2$ and $s=1$,  or,   $\pi=3$   and   $1\leq s\leq 5$.

case 2: $\pi\mid pq$:

Since further $p$ and $q$ are coprime, there are exactly two possibilities:

sub-case 1: $\pi\mid p$ and $\pi\nmid q$:

From $\pi\mid p$ and $\pi\mid A=p(59p^{2}-236q)-7q^{2}$, we deduce that $\pi\mid 7q^{2}$. As further $\pi\nmid q$, then, $\pi=7$. In particular, $7\nmid q$. Now, $\pi^{s}=7^{s}$ divides $2A+B=-2\cdot 3^{5}\cdot pq$, where $7^{s}$ is coprime to $2\cdot 3^{5}\cdot q$. By Gauss’ theorem, $7^{s}$ divides $p$. Finally, from $7^{s}\mid p$ and $7^{s}\mid A=p(59p-236q)-7q^{2}$, we deduce that $7^{s}\mid 7q^{2}$. Since further $7^{s}$ is coprime to $q^{2}$, we get, $7^{s}\mid 7$. Hence $s=1$ (so $\pi^{s}=7^{1}$).

sub-case 2: $\pi\mid q$ and $\pi\nmid p$:

From $\pi\mid q$   and   $\pi\mid A=q(-236p-7q)+59p^{2}$, we deduce that $\pi\mid 59p^{2}$. As further $\pi\nmid p$, then, $\pi=59$. In particular, $59\nmid p$. Now, $\pi^{s}=59^{s}$ divides $2A+B=-2\cdot 3\cdot pq$, where $59^{s}$ is coprime to $2\cdot 3^{5}\cdot p$.   By Gauss’ theorem, $59^{s}$ divides $q$.   Finally,  from   $59^{s}\mid q$   and

$59^{s}\mid A=q(-236p-7q^{2})+59p^{2}$,   we deduce that $59^{s}\mid 59p^{2}$.

Since further $59^{s}$ is coprime to $p^{2}$, we get $59^{s}\mid 59$.   Hence $s=1$ (so $\pi^{s}=59^{1}$). 2. (ii)

1. $\bullet$

If $7\mid\delta$, then $7\mid B=7(-2pq+2q^{2})-118p^{2}$. Hence, $7\mid 118p^{2}$. As further $7\nmid 118$, then,

$7$ must divide $p$.

Conversely, if $7\mid p$, then, $7\mid A=p(59p-236q)-7q^{2}$  and   $7\mid B=p(-118p-14q)+7(2q^{2})$. Hence, $7\mid\delta$. 2. $\bullet$

Similarly, if $59\mid\delta$, then $59\mid A=59(p^{2}-4pq)-7q^{2}$. Hence, $59\mid 7q^{2}$. Then, clearly, $59\mid q$. Conversely, if $59\mid q$, then $59\mid A=59p^{2}+q(-236p-7q)$ and $59\mid B=59(-2p^{2})+q(-14p+14q)$. Hence, $59\mid\delta$.

$\square$

Next, we give a parametrization of the rational points on the ellipse $7X^{2}+59Y^{2}=3$: Given $p,q\in\mathbb{Z}^{\ast}$, set $A=A(p,q)$, $B=B(p,q)$ and $C=C(p,q)$. In virtue of lemma $1$ $(i)$, we have $7A^{2}+59B^{2}=3C^{2}$. Since $C>0$, we obtain $7\left(\frac{A}{C}\right)^{2}+59\left(\frac{B}{C}\right)^{2}=3$. Hence, $(X,Y)=\left(\frac{A}{C},\frac{B}{C}\right)$ is a rational point on the ellipse $7X^{2}+59Y^{2}=3$.

Now, we focus with more attention and precision on the converse:

Proposition 2:

Let $(X,Y)$ be a rational point on the ellipse $7X^{2}+59Y^{2}=3$, such that $X\neq\frac{1}{9}$ and $Y\neq\frac{2}{9}$. Then, we have

[TABLE]

for some $p,q\in\mathbb{Z}^{\ast}$,   $q>0$ ,   $p$ and $q$ coprime, and where the fraction (in lowest terms) $\frac{p}{q}$ is precisely equal to the non-zero rational number $\theta=\frac{9Y-2}{9X-1}$.

[TABLE]

Using $(4.1)$, we then have,    $7X^{2}+59\left(\theta X+\frac{2-\theta}{9}\right)^{2}=3$.   Multiplying by $81$ yields

[TABLE]

Expanding $(4.2)$ and after a little algebra, we find

[TABLE]

The reduced discriminant of this trinomial in $X$ is $\Delta=81(118\theta+7)^{2}$. Hence, $\sqrt{\Delta}=9|118\theta+7|$, so,

[TABLE]

According to the hypothesis, the root $X=\frac{1}{9}$   has to be rejected.   Therefore, we must have

[TABLE]

Using $(4.1)$ and $(4.3)$, we obtain

[TABLE]

Finally, let us write $\theta$ in lowest terms as $\theta=\frac{p}{q}$, where $p,q\in\mathbb{Z}^{\ast}$, $q>0$ ($p$ and $q$ coprime).

Replacing $\theta$ by $\frac{p}{q}$ in $(4.3)$ and $(4.4)$ yields

[TABLE]

and

[TABLE]

∎

$\bullet$ Finally, we study the diophantine equation

[TABLE]

where the variables $x,y,z$ are non-zero integers, and where $z>0$. A solution of $(2)$ is said to be positive when furthermore $x$ and $y$ are positive. A solution $(x,y,z)$ of $(2)$ is called primitive when $x$ and $y$ are coprime (equivalently, when $x,y,z$ are pairwise coprime).

Notation:

A "pp-solution" of $(2)$ stands for: a primitive and positive solution of $(2)$.

Note first that if $p,q\in\mathbb{Z}^{\ast}$, and if $x=A(p,q)$, $y=B(p,q)$ and $z=C(p,q)$, then, due to lemma $1(i)$, $(x,y,z)$ is a solution of $(2)$. However, even if $p$ and $q$ are coprime, this solution is not necessarily primitive.

We are particularly interested in the converse, that we study with more attention and precision:

Definition:

Let $S=(x,y,z)$ be a primitive solution of $(2)$, with $z>0$, such that,

[TABLE]

We define the incidence of $S$, that we denote by $I(S)$, as the non-zero rational number

[TABLE]

Clarification:

We clarify why $I(S)$ is well-defined and why $I(S)\neq 0$:

If we suppose that $9x-z=0$, then we get $\frac{x}{z}=\frac{1}{9}$. Since the fractions $\frac{x}{z}$ and $\frac{1}{9}$ are both in lowest terms, and since $z>0$, we obtain $z=9$ and $x=1$. From this and $7x^{2}+59y^{2}=3z^{2}$, we get $y=\pm 2$, so $(x,y,z)=(1,\pm 2,9)$, a contradiction.

On the other hand, if we suppose that $9y-2z=0$, then we get $\frac{y}{z}=\frac{2}{9}$. Since both fractions $\frac{y}{z}$ and $\frac{2}{9}$ are in lowest terms, and since $z>0$, we obtain $z=9$ and $y=2$. From this and $7x^{2}+59y^{2}=3z^{2}$, we get $x=\pm 1$, so $(x,y,z)=(\pm 1,2,9)$, a contradiction.

**Proposition 3:

**Let $S=(x,y,z)$ be a primitive solution of $(2)$, with $z>0$, such that

[TABLE]

Write the incidence $I(S)$ in lowest terms as $I(S)=\frac{p}{q}$, with $q>0$. Set $A=A(p,q)$, $B=B(p,q)$, $C=C(p,q)$, and let $\delta$ be the positive gcd of $A$ and $B$. Then, we have precisely:

[TABLE]

where $\delta$ divides $2\cdot 3^{5}\cdot 7\cdot 59$.

Furthermore, we have     [ $7\mid\delta\iff 7\mid p$ ]     and     [ $59\mid\delta\iff 59\mid q$ ].

[TABLE]

$\bullet$ Now,

[TABLE]

Hence, $I(S)=\frac{p}{q}$ (where $\frac{p}{q}$ is in lowest terms, with $q>0$). Next, let $\delta$ be the positive gcd of $A$ and $B$. By proposition $1$, $\delta$ divides $2\cdot 3^{5}\cdot 7\cdot 59$, where we have $[\ 7\mid\delta\iff 7\mid p\ ]$ and $[\ 59\mid\delta\iff 59\mid q\ ]$.

Finally, we have $X=\frac{A}{C}$, hence $\frac{x}{z}=\frac{A/\delta}{C/\delta}$, where both fractions are in lowest terms with $z>0$ and $\frac{C}{\delta}>0$. We conclude that $x=\frac{A}{\delta}$ and $z=\frac{C}{\delta}$.

Similarly, we have $Y=\frac{B}{C}$, hence, $\frac{y}{z}=\frac{B/\delta}{C/\delta}$, where both fractions are in lowest terms with $z>0$ and $\frac{C}{\delta}>0$. We conclude that $y=\frac{B}{\delta}$ (and $z=\frac{C}{\delta}$). ∎

6. Proof of theorem 2:

Lemma 6:

Let $(x,y,z)$ be a pp-solution of $(2)$, with $z\neq 9$ ($z>0$). Note that the incidences of the $4$ primitive solutions of $(2)$, $(\pm x,\pm y,z)$, are all defined. Then, among these $4$ solutions, there is (at least) one, say $S=(x^{\prime},y^{\prime},z)$ (where $x^{\prime}\in\{\pm x\}$ and $y^{\prime}\in\{\pm y\}$), such that, if $I(S)$ is written in lowest terms as $I(S)=\frac{p}{q}$, with $q>0$, and, if $A=A(p,q)$, $B=B(p,q)$ and if $\delta$ is the positive gcd of $A$ and $B$, then, $\delta$ is neither divisible by $7$ nor by $59$.

[TABLE]

Since $z\neq 9$, then, each of these solutions is $\neq(\pm 1,\pm 2,9)$, so that its incidence is defined. Let us write $I(S_{i})$, $i=1,2,3,4$, in lowest terms as $I(S_{i})=\frac{p_{i}}{q_{i}}$, with $q_{i}>0$.   We have precisely

[TABLE]

Since the fractions $\frac{p_{i}}{q_{i}}$ are irreducible, there are $4$ integers $\lambda_{i}$ such that

                  $9y-2z=\lambda_{1}p_{1}\ \ \ \ (1,a)$ $9x-z=\lambda_{1}q_{1}\ \ \ \ (1,b)$                  

                $-9y+2z=\lambda_{2}p_{2}\ \ \ \ (2,a)$ $9x+z=\lambda_{2}q_{2}\ \ \ \ (2,b)$                  

                  $9y+2z=\lambda_{3}p_{3}\ \ \ \ (3,a)$ $-9x+z=\lambda_{3}q_{3}\ \ \ \ (3,b)$                  

                  $9y+2z=\lambda_{4}p_{4}\ \ \ \ (4,a)$ $9x+z=\lambda_{4}q_{4}\ \ \ \ (4,b)$                  

For $i=1,2,3,4$, set $A_{i}=A(p_{i},q_{i})$, $B_{i}=B(p_{i},q_{i})$ and let $\delta_{i}$ be the positive gcd of $A_{i}$ and $B_{i}$. Recall (by proposition $1(ii)$) that, for $i=1,2,3,4$, we have:

[TABLE]

Now, let us focus on the following arguments $(A1)$ and $(A2)$:

(A1) If $7$ were to divide $p_{1}$ and $p_{3}$, then, by $(1,a)$ and $(3,a)$, $7$ would divide $(9y-2z)\pm(9y+2z)$, that is, $7$ would divide $18y$ and $-4z$, hence, $7$ would divide $y$ and $z$, a contradiction since $\gcd(y,z)=1$. Similarly, if $7$ were to divide $p_{1}$ and $p_{4}$, then using $(1,a)$ and $(4,a)$, we would obtain the same contradiction.

$\bullet$ Argument $(A1)$ shows that:

[TABLE]

(A2) If $7$ were to divide $p_{2}$ and $p_{3}$, then, by $(2,a)$ and $(3,a)$, $7$ would divide $(-9y+2z)\pm(9y+2z)$, that is, $7$ would divide $4z$ and $-18y$, hence, $7$ would divide $z$ and $y$, a contradiction since $\gcd(y,z)=1$. Similarly, if $7$ were to divide $p_{2}$ and $p_{4}$, then using $(2,a)$ and $(4,a)$, we would obtain the same contradiction.

$\bullet$ Argument $(A2)$ shows that:

[TABLE]

From arguments $(A1)$ and $(A2)$, we deduce that:

[TABLE]

Since [$\,7\mid p_{i}\iff 7\mid\delta_{i}\,$], we conclude the following:

[TABLE]

Next, we provide two similar arguments $(A3)$ and $(A4)$, relative to the prime $59$:

(A3) If $59$ were to divide $q_{1}$ and $q_{2}$, then, by $(1,b)$ and $(2,b)$, $59$ would divide $(9x-z)\pm(9x+z)$, that is, $59$ would divide $18x$ and $-2z$, hence, $59$ would divide $x$ and $z$, a contradiction since $\gcd(x,z)=1$.

Similarly, if $59$ were to divide $q_{1}$ and $q_{4}$, then, using $(1,b)$ and $(4,b)$, we would obtain the same contradiction.

$\bullet$ Argument $(A3)$ shows that:

[TABLE]

(A4) If $59$ were to divide $q_{3}$ and $q_{2}$, then, by $(3,b)$ and $(2,b)$, $59$ would divide $(-9x+z)\pm(9x+z)$, that is, $59$ would divide $2z$ and $-18x$, hence, $59$ would divide $z$ and $x$, a contradiction, since $\gcd(x,z)=1$. Similarly, if $59$ were to divide $q_{3}$ and $q_{4}$, then using $(3,b)$ and $(4,b)$, we would obtain the same contradiction.

$\bullet$ Argument $(A4)$ shows that:

[TABLE]

From arguments $(A3)$ and $(A4)$, we deduce that:

[TABLE]

Since [$\,59\mid q_{i}\iff 59\mid\delta_{i}\,$], we conclude the following:

[TABLE]

Finally, by combining $(R1)$ and $(R_{2})$ (that leads to $4$ cases), we see that for some $i\in\{1,2,3,4\}$,

$\delta_{i}$ is neither divisible by $7$ nor by $59$. ∎

Now we are in a position to prove the crucial result from which we will deduce theorem 2.

**Lemma 7:

**Let $S_{0}=(x,y,m)$ be a pp-solution of $(1)$, such that $S_{0}\neq(1,2,5)$. Then, $S_{0}$ is one of the (two) successors of some smaller pp-solution of $(1)$.

[TABLE]

As further $\gcd(x,y)=1$, we see that $(x,y,z)$ is a pp-solution of $(2)$. Since $z\geq 3^{7}>9$, then, by lemma $6$, there are $x^{\prime}\in\{\pm x\}$ and $y^{\prime}\in\{\pm y\}$ such that, with $S=(x^{\prime},y^{\prime},z)$, if $I(S)$ is written in lowest terms as $I(S)=\frac{p}{q}$, with $q>0$, if $A=A(p,q)$, $B=B(p,q)$, $C=C(p,q)$, and if $\delta$ is the positive gcd of $A$ and $B$, then $\delta$ is neither divisible by $7$ nor by $59$. But, by proposition $1$, $\delta$ divides $2\cdot 3^{5}\cdot 7\cdot 59$. We conclude here that $\delta$ must divide $2\cdot 3^{5}$. In such situation, we claim that $\delta$ cannot have the factor $2$: Indeed, for the purpose of contradiction, suppose that $\delta=2\cdot 3^{s}$, $0\leq s\leq 5$. By proposition $3$, we would get $z=\frac{C}{\delta}$. Hence, $C=\delta z=2\cdot 3^{s}\cdot 3^{n}=2\cdot 3^{n+s}$. That is, $9(7q^{2}+59p^{2})=2\cdot 3^{n+s}$. Therefore, we would get

[TABLE]

(where $n+s-2\geq 7+0-2=5$).

However, this is a contradiction, according to corollary $4$.   Finally, $\delta$ must have the form

$\delta=3^{s}$, $0\leq s\leq 5$.  Recall by proposition $3$, that

[TABLE]

We consider two cases:

Case 1: $\delta=1$

We then have $x^{\prime}=A(p,q)$,  $y^{\prime}=B(p,q)$ and $z=C(p,q)$.

From $z=3^{n}=C(p,q)=9(7q^{2}+59p^{2})$, we obtain

[TABLE]

Therefore, $(q,|p|,n-2)$ is a pp-solution of $(1)$. Now, we have

[TABLE]

and

[TABLE]

Hence, in virtue of remark $3$ $(i)$, we conclude that

[TABLE]

(Note that $2n+1>n-2$).

Case 2: $3\mid\delta$

In particular, $3\mid A$.   If we had $pq\equiv 1\pmod{3}$, we would get

[TABLE]

which is in contradiction with $3\mid A$. Hence, $pq\equiv-1\pmod{3}$. By corollary $2$, we get $\delta=3^{5}$.

Hence, $x^{\prime}=3^{-5}A$,   $y^{\prime}=3^{-5}B$ and $z=3^{-5}C$. We may write

[TABLE]

Hence,

[TABLE]

Therefore, $(q,\ |p|,\ n+3)$ is a pp-solution of $(1)$.   Now, we have

[TABLE]

and

[TABLE]

Hence, in virtue of remark $3$ $(ii)$, we conclude that

[TABLE]

(As $n\geq 7$, then, $2n+1>n+3$). ∎

Proof of theorem 2:

Recall that $S_{1}=(1,2,5)$ is the smallest pp-solution of $(1)$.

Let $T_{1}=(x_{1},y_{1},m_{1})$ be a pp-solution of $(1)$. If $T_{1}=S_{1}$, there is nothing to prove. From now on, we assume that $T_{1}\neq(1,2,5)$.

By lemma $7$, $T_{1}$ is a successor of a smaller pp-solution of $(1)$, say

[TABLE]

If $T_{2}=(1,2,5)$, we stop the process. Otherwise, by lemma $7$, $T_{2}$ is a successor of a smaller pp-solution of $(1)$, say

[TABLE]

This process must stop, otherwise we would obtain an infinite strictly decreasing sequence of positive integers $m_{1}>m_{2}>m_{3}>\cdots$, which is impossible !

Hence, for some $r\geq 2$, $T_{r}=(x_{r},y_{r},m_{r})=(1,2,5)$. In other words, $T_{1}$ is a successor of a successor $\ldots$ of a successor of $S_{1}$,   so that $T_{1}$ belongs to the binary tree described in corollary $1$. $\square$

**Remark 5:

**A direct consequence of theorem $2$ is the uniqueness of a pp-solution $(x,y,m_{0})$ of $(1)$, for a given suitable $m_{0}=10k_{0}+5$.

$\bullet$ Finally, noting that in any solution $(x,y,m)$ of $(1)$, $x$ and $y$ have opposite parities we show that the case $[\,x$ even and $y$ odd$\,]$ never arises.

**Proposition 4:

**Let $(x,y,m)$ be any solution (not necessarily primitive) of $(1)$. Then, $x$ is odd and $y$ is even.

By theorem $2$, the pp-solutions of $(1)$ lie all on the binary tree of the iterated successors of $S_{1}$.

We proceed by induction on the binary tree:

The property holds for $S_{1}$.

Next, we assume that the property holds for a pp-solution $S=(q,p,\omega)$ of $(1)$, and we prove that the property holds for the two successors of $S$. Let $S^{\prime}=(x^{\prime},y^{\prime},m^{\prime})$ and $S^{\prime\prime}=(x^{\prime\prime},y^{\prime\prime},m^{\prime\prime})$ denote respectively the first and second successor of $S$.

Since $B(p,q)=-118p^{2}-14pq+14q^{2}$ is even, and $y^{\prime}=3^{-5}|B(p,q)|$, then $y^{\prime}$ is even (so $x^{\prime}$ is odd). (Alternatively, from $q$ odd and $p$ even, we see that $A(p,q)$ is odd, so $x^{\prime}=3^{-5}|A(p,q)|$ is odd).

Similarly,  since  $B(-p,q)=-118p^{2}+14pq+14q^{2}$ is even, and $y^{\prime\prime}=|B(-p,q)|$, then $y^{\prime\prime}$ is even

(so $x^{\prime\prime}$ is odd). ∎

Exercises:

The equation $3x^{2}+5y^{2}=7^{m}$ has no solution in integers $x,y,m$. 2. 2.

In any integer solution of $2x^{2}+3y^{2}=5^{m}$, $m$ must be odd. 3. 3.

$(\star)$ Can one find distinct positive primes $a,b,c$ such that the equation $ax^{2}+by^{2}=c^{z}$ has exactly one primitive and positive solution? 4. 4.

$(\star)$ Let $(x,y,m)$ be a positive and primitive solution of $7x^{2}+59y^{2}=3^{m}$. Prove or disprove that $xy\equiv-1\pmod{3}$.

References

1. [1]

W. Sierpinsky, On the equation $3^{x}+4^{y}=5^{z}$, Wiadom. Mat. (2), 1955/1956,  1  194-195

(in Polish). 2. [2]

N. Terai, The Diophantine Equation $a^{x}+b^{y}=c^{z}$,  Proc. Japan Acad., 1994, 70A(2)  22-26. 3. [3]

Bonyat Sroysang, More on the Diophantine Equation $8^{x}+19^{y}=z^{2}$, International Journal of Pure and Applied Mathematics, Vol. 81,  No. 4,  2012,   601-604. 4. [4]

Julius Fergy T. Rabago, On the Diophantine Equation $2^{x}+17^{y}=z^{2}$, J. Indones. Math Soc., Vol. 22, No. 2 (2016)  pp. 85-88. 5. [5]

Zhenfu Cao, Chuan I Chu and Wai Chee Shiu, *The Exponential Diophantine Equation

$AX^{2}+BY^{2}=\lambda k^{z}$ and its Applications*, Taiwanese Journal of Mathematics, Vol. 12, No. 5, 2008,  1015-1034. 6. [6]

Borevitch Z.I., Shafarevich I.R., Number theory, Acad. Press, 1966, Chap. 1, Par. 7. 7. [7]

Niven I., Zuckerman H.S., An Introduction to the theory of Numbers, 4th Ed., John Wiley

& Sons. 8. [8]

Hardy G.H., Wright E.M., An Introduction to the Theory of Numbers, Oxford, 5th Ed., 1979. 9. [9]

Tijdeman R., Exponential Diophantine Equations, Proceedings of the International Congress of Mathematicians, Halsinki (381-387), 1978. 10. [10]

Andreescu T., Andrica D., Number Theory, Structures, Examples, and Problems, Birkhäuser, Boston-Basel-Berlin, 2009. 11. [11]

MORDELL, L.J., Diophantine Equations, Acad. Press, London, 1969. 12. [12]

Samuel Pierre, Théorie Algébrique des Nombres, Hermann, 1967.