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$$3^x = 2+y^2$$

Solve for the integer values of $x$ and $y$. Can this be solved using graph theory and calculus? I am trying out for half an hour and uses most of my preknowledge but still it is not solved, however I found two solutions for $(x,y)$ which are $(1,1)$ and $(3,5)$.

J. W. Tanner
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aman rana
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    Please include your thoughts and efforts on the problem. This helps to give an answer more suited to your needs. – Servaes Feb 04 '19 at 11:13
  • No, a solution with calculus or a grpahical solution is not possible. Besides $(1/\pm 1)$ and $(3/\pm 5)$ there are probably no integers solutions. Upto $x=10^5$, there is no further solution. Probably someone on this site will find a complete solution, I admit I have no idea for a proof. – Peter Feb 04 '19 at 11:23
  • First thoughts: Taking mod 3 on both sides, $y$ has to be congruent to either 1 or 2 (mod 3) – Matti P. Feb 04 '19 at 11:25

1 Answers1

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Let $x$ and $y$ be integers such that $3^x=2+y^2$. Note that $2+y^2>1$ and hence $x>0$.

In the ring $\Bbb{Z}[\sqrt{-2}]$ the element $3$ factors as $3=(1+\sqrt{-2})(1-\sqrt{-2})$, and we have $$3^x=(y+\sqrt{-2})(y-\sqrt{-2}).$$ The gcd of the two factors on the right hand side divides $$(y+\sqrt{-2})-(y-\sqrt{-2})=2\sqrt{-2}=-(\sqrt{-2})^3,$$ and it also divides $3^x$, and so the two factors are coprime. Because $\Bbb{Z}[\sqrt{-2}]$ is a unique factorization domain, it follows that $y+\sqrt{-2}=(1+\sqrt{-2})^x$ and $y-\sqrt{-2}=(1-\sqrt{-2})^x$.

I'll leave it for you (for now) to check that the coefficient of $\sqrt{-2}$ in the expansion of $(1+\sqrt{-2})^x$ equals $1$ if and only if $x=1$ or $x=3$.

Servaes
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