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I believe there are only two non-negative integer solutions to $$2^{2a} + 3^{2b} = 5^c.$$ The solutions I have are $a=1,b=0,c=1$ and $a=2,b=1,c=2$. I'm not certain this is correct. I'd like to know if there are other solutions, or if there a theorem that says these are the only solutions.

For further context, I'm generating Pythagorean triples as $$(r^2 - s^2, 2rs, r^2 + s^2)$$ where $r$ and $s$ are powers of $2$ or $3$. The solutions above correspond to the $(3,4,5)$ and $(7,24,25)$ Pythagorean triples.

Thanks for reading,

Daniel

Daniel Mansfield
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  • How do you know that $c$ = an odd positive integer won't generate any solutions? – user2661923 Apr 29 '21 at 00:37
  • I'm not sure what you mean. c=1 does correspond to a solution. I should have added that I can prove this myself when c is even, so it really is just the c="odd positive integer > 1" that I can't resolve. – Daniel Mansfield Apr 29 '21 at 00:42
  • Sorry, my bad. I intended $c = $ an odd positive integer $> 1.$ Also, assuming that $c = 2d,$ then you have that $2^{(2a)} + 3^{(2b)} = 5^{(2d)} \implies$ that you do have a primitive pythagorean triplet $(x^2 + y^2 = z^2, ~~~x,y,z~$ pairwise comprime $)$. This means that your $r,s$ specification holds. I would like to see a proof that when $c$ is even, the only solutions are $(3,4,5), (7,24,25).$ If you can, please edit your query to show this proof. – user2661923 Apr 29 '21 at 00:54
  • Re my last comment, my bad again. $(7,24,25)$ won't be a solution, so the partial-assertion is that when $c$ is even, the only solution is $(3,4,5)$. – user2661923 Apr 29 '21 at 01:18
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    Does this answer your question? Positive integral solutions of $3^x+4^y=5^z$. Note $2^{2a} = 4^{a}$ so your equation could be written as $3^{2b} + 4^{a} = 5^{c}$. Also, although the proposed duplicate doesn't include the $b = 0$ solution, it otherwise actually answers a somewhat more general question than yours. – John Omielan Apr 29 '21 at 01:23
  • If $c=2d$ is even then $3^{2b}=5^{2d}-2^{2a}=(5^d-2^a)(5^d+2^a)$ so both factors on the right must be powers of $3$, but the gcd of the two terms on the right divides $2\times 5^d$ and $2^{a+1}$. – lulu Apr 29 '21 at 01:28
  • @JohnOmielan yes that will work provided we also assume the integers $x,y,z$ are positive. Thanks for the recommendation! – Daniel Mansfield Apr 29 '21 at 04:34
  • @user2661923 The solution suggested by John is more or less what I did. All I missed was that the power of $5$ had to be even. – Daniel Mansfield Apr 29 '21 at 04:35

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