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What are all the solutions to

$$2^{2x}+2^{2y}+1=n^2 $$

I tried using the parametrization of Pythagorean Quadruples, but it did not work quite well.

There are $2$ parametrizations:

$(2np,2mp,p^2-n^2-m^2,p^2+n^2+m^2)$

Or $(mp+nq,np-mq,p^2+q^2-n^2-m^2,p^2+q^2+n^2+m^2)$

The problem is that the first one doesn't generate all solutions, but worked in my proof.So...any help?

EDIT:What we want to show is that the only solutions are

${(2^{2y-1}+1)}^2=4^{2y-1}+4^y+1$

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    $\begingroup$ Do all the solutions have to be integers? $\endgroup$ – Peter Woolfitt Jan 29 '15 at 18:32
  • $\begingroup$ Why Pythagorean Triples? Is it a hint to this question? $\endgroup$ – agha Jan 29 '15 at 18:37
  • $\begingroup$ One solution would be $x=y=1$ $n=3$. $\endgroup$ – ghosts_in_the_code Jan 29 '15 at 18:41
  • $\begingroup$ Sorry about that, I meant Pythagorean Quadruples $\endgroup$ – user211570 Jan 29 '15 at 18:46
  • $\begingroup$ Source of the problem??? $\endgroup$ – Will Jagy Jan 29 '15 at 19:09
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We must have $$ 4^x + 4^y = (n-1)(n+1) \tag{1}$$ and $\gcd(n-1,n+1)\leq 2$, so by assuming $x\geq y$ and rewriting $(1)$ as $$ 4^y\cdot\left(4^{x-y}+1\right) = (n-1)(n+1) \tag{2}$$ there are just a few cases to check.

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  • $\begingroup$ Tried that, but didnt go so well.Could you elaborate how you manipulated the cases? $\endgroup$ – user211570 Jan 29 '15 at 18:44
  • $\begingroup$ By the way, you are on brilliant, arent you? $\endgroup$ – user211570 Jan 29 '15 at 18:47
  • $\begingroup$ If $n$ is even, we must have $y=0$ and $4^x+2=n^2$, no hope. If $x-y\geq 1$ and $n$ is odd, we must have $(n\pm 1)=2(4^{x-y}+1)$ and $(n\mp 1)=2^{2y-1}$. If $x-y=0$, we must have $2\cdot 4^x+1 = n^2$, so $x=1$ and $n=3$. $\endgroup$ – Jack D'Aurizio Jan 29 '15 at 18:51
  • $\begingroup$ But why cant, for instance, n+1=2.a and n-1=2^{2y-1}b, where a.b=4^{x-y}+1? $\endgroup$ – user211570 Jan 29 '15 at 18:53
  • $\begingroup$ @user211570: oh, ok, that may happen. $\endgroup$ – Jack D'Aurizio Jan 29 '15 at 18:57

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