# Solution of Diophantine equation

I have seen $3^x$ + $3^y$ = $6^z$ and $4^x$ + $18^y$ = $22^z$ on lecture series of Prof. Gandhi. In my own study, I have constructed the following theorem (I am not sure about solvability) and I am seeking some good discussion on this theorem including proof/ comments etc.

If A = {(2k-1, 2k-1, $2^k$)|k is a positive integer}, B = {(2k+3, 2k, 3.$2^k$)|k is a non negative integer} and C = {2k, 2k+3, 3.$2^k$)|k is a non negative integer}, then the solution of $2^x$ + $2^y$ = $z^2$ is (x, y, z) belongs to $A \cup B \cup C$.

Also discuss that, if P is prime and > 2 then solution set of $2^x$ + $p^y$ = $z^2$ will be...

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In general it's a bad idea to just simply say "discuss this"; ask a concrete question like "is this true / how do we show this?" Also, I think you mean $3\cdot2^k$ rather than $3.2^k$ and the cartesian product $A\times B\times C$ rather than the set union $A\cup B\cup C=\mathbb{Z}_+$. – anon Nov 5 '11 at 8:57
Thank you and I will use your suggestions in future posts – mathew Nov 6 '11 at 17:35

I think you are asking for the solutions to $$2^x+2^y=z^2$$ If $x$ and $y$ are both positive then $z$ must be even and you can divide out by powers of 2 until you get to $$1+2^r=s^2$$ (unless $x=y$, which case is easy to handle) which is $$2^r=(s+1)(s-1)$$ which tells you $s+1$ and $s-1$ must both be powers of 2. That will only happen for $s=3$, and thus $r=3$. Working your way back, you get to the solutions you have found.

EDIT: Now let's look at $$2^x+p^y=z^2$$ with $p\gt2$ prime. If $x=2r$ is even, then $p^y=(z+2^r)(z-2^r)$, so both $z+2^r$ and $z-2^r$ are powers of $p$, so the difference, $2^{r+1}$, is divisible by $p$, contradiction. Well, unless $z-2^r=1$, $z+2^r=p^y$, so $p^y=2^{r+1}+1$. For $r=0,1,3,7,15$ we get $1+3=2^2$, $4+5=3^2$, $64+17=9^2$, $2^{14}+257=129^2$, $2^{30}+65537=(2^{15}+1)^2$. But to get all the solutions, you would have to know all the Fermat primes, and more, and that's not likely to happen anytime soon.

If $x$ is odd and $y=1$ then you are asking for $z^2-Q$ to be a prime, where $Q$ is an odd power of 2, and everyone believes that happens infinitely often, and no one can prove it, and there's no hope of finding any formulas for it. So I think you are asking for too much.

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Can you discuss the last part of my post little bit more? – mathew Nov 6 '11 at 17:38