Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

Thanks in advance.

share|improve this question
5  
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

1 Answer 1

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.