finding A using with restriction $1 \leq a \leq 20$ in GCD

Question

For what $1 \leq a \leq 20$ you are finding $a$ is it true that $a^m+a^n=x^2$ for positive integers $a,m,n,x.$

I did

$a^m+a^n=x^2.$

$=a^m(a^{n-m}+1)=x^2$

We know that since $(a,b)=1$ since the GCD is $1,$ we can say

$= (a^m,a^k+1)=1$

I know that $a^m$ is prime which is the only factor of $a.$ $a$ is not in the factor for $a^k+1.$

I am not sure how you would solve this problem with this. Can someone help me with this?

Answer 1

Since $\gcd(a^m, a^k+1) = 1$, both $a^m$ and $a^k+1$ must be squares.

If $k=1$, then given $1\le a\le 20$ we must have $a\in\{3,8,15\}$. Since none of these possible values of $a$ is itself a square, we must have $m=2r$ even, so we get the three families of solutions

• $3^{2r}+3^{2r+1} = (2\cdot 3^r)^2$
• $15^{2r}+15^{2r+1} = (4\cdot 15^r)^2$.
• $8^{2r}+8^{2r+1} = (3\cdot 8^r)^2$.

Alternatively, if $k>1$ then using Catalan's conjecture, we see that we must have $a=2$, $k=3$, and $a^k+1 = 9$. Since $a^m = 2^m$ must be a square, we must have $m$ even. So the only solutions in this case are given by $a=2$, $m=2r$ even, $n = m+3$:

• $2^{2r} + 2^{2r+3} = (3\cdot 2^r)^2$.

Finally, note that the solutions for $a=8$ are a subset of those for $a=2$, so you end up with just three families of solutions, for $a=2$, $a=3$, and $a=15$.