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:


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


  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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