Square Numbers general methods

Question

Let this equation be given: $$a^x+a^y=b^2$$ With which method can one determine if there are infinitely many or no solutions for a random number $a$ so that $b^2$ is a square number.

$a, b, x$ and $y$ have to be positive whole numbers.

Answer 1

A fuller exposition of solutions to $a^x+a^y=b^2$

WLOG, let $x\le y$ $$a^x+a^y=b^2 \Rightarrow a^x(a^{y-x}+1)=b^2$$ Now $\gcd(a^x,a^{y-x}+1)=1$, and since $b^2$ is a perfect square, its relatively prime factors $a^x$ and $a^{y-x}+1$ must each be perfect squares. So the exponent $x=2n$. This further means for some integer $c$, $c^2-(a^{y-x})=1$. However, Catalan's conjecture establishes that if $y-x>1$, the only positive integer solution to that equation is $c=3,a=2,(y-x)=3$. This affords the first family of solutions: $$2^{2n}+2^{2n+3}=2^{2n}(8+1)=b^2 \\ (3\cdot 2^n)^2=b^2 \\ a=2,b=(3\cdot 2^n),x=2n,y=2n+3$$ Examples in this family include $4+32=36,\ 16+128=144$

Next, we look at the case $x=y,\ (y-x)=0$. This corresponds to $2a^x=b^2$ implying first $2\mid b$ and next that $a^x$ must have a factor of $2$ to an odd exponent. That further requires that $x$ be odd, and that $a$ must have a factor of $2$ to an odd exponent. Finally, any odd prime factor of $a$ must be present an even number of times. To recap: $x=y=(2k-1); \ a=2^{2n-1}c^{2m}$ where $c$ is some odd integer. This affords a second family of solution: $$2\cdot (2^{2n-1}c^{2m})^{2k-1}=b^2$$ Examples in this family include $8+8=16, \ 18^3+18^3=108^2,\ 800^5+800^5=25600000^2$

Finally we look at the remaining case $y-x=1$. This requires $a^{2n}(a+1)=b^2$. Plainly, this holds whenever $a=d^2-1$, yielding $b=d(d^2-1)^n$

Examples in this family include $48^5+48^4=16128^2,\ 168^3+168^2=2184^2$

Choices of $a$ other than $2,\ 2^{2n-1}c^{2m},\ d^2-1$ will fail to afford solutions.

Answer 2

My try:

Let $x=1, y=1$, and you have $2a=b^2$, where $a=8,b=4$.

Let $a=7$, and $a^x$ and $a^y \equiv 1$ or $3 \pmod {4}$. Which means they'll sum to give $a^x+a^y \equiv 0$ or $2 \pmod{4}$. Since any square is $\equiv 0$ or $1 \pmod{4}$, we'd need the $0$ in both cases. To get that, $x+y$ would need to be odd.

But $$7^x+7^y=7^x+7^x*7^{y-x}=7^x(1+7^{y-x})$$ will give a factor of two with an odd exponent, meaning the whole thing can't be square.