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For integers $x, y>0$ solve the following Equation: $$x^x+1=y^2$$

Well, $x$ must be odd. Because for $x=2z$ we get difference of two squares as $1$, which is impossible in positive integers. We have $$x^x=(y-1)(y+1).$$ Thus for every prime divisor of $x$ like $p$, if $v_p(x)=k$, then $y=p^{kx}z \pm 1$. Where $\gcd(p, z)=1$. I am not sure how to use this! Any ideas?

Edit: Mihăilescu's theorem will do this easily. I am interested in an alternative elementary method. If you can think of one please let me know.

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    $\begingroup$ C a t a l a n . $\endgroup$ – franz lemmermeyer Jun 17 '17 at 20:33
  • $\begingroup$ K i l l e d i t ! $\endgroup$ – Ghartal Jun 17 '17 at 20:41
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    $\begingroup$ If $x$ is odd, $y$ is even and so $y-1,y+1$ are relatively prime, hence must both be $x$th powers... $\endgroup$ – Wojowu Jun 17 '17 at 20:50
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    $\begingroup$ And if $1=y^2-(2u)^{2u}=(y-(2u)^u)(y+(2u)^u)$.... $\endgroup$ – Barry Cipra Jun 17 '17 at 20:53
  • $\begingroup$ @Wojowu very good point. Thanks. $\endgroup$ – Ghartal Jun 17 '17 at 21:06
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We can immediately exclude $x=1$ and $x=0$ , the latter because it is standard that $0^0=1$ is assumed.


Assume $x=2w$ being even. Then we get $$ y^2 - (2w)^{2w}= 1 \\ (y-x^w)(y+x^w) = 1 \\ \to x^w=(2w)^w =0,y=1 $$ There is no $w$ allowing $(2w)^w=0$ , thus no solution. The same answer that you've already found.
Now assume $x=2w+1$ being odd. We rewrite $$ x^x = y^2-1 = (y-1)(y+1)$$

  • Assume $y=2z$ being even. Then $x^x$ must have two coprime odd factors, so $x=pq$ Then $$ (y-1)(y+1)=(pq)^x = p^x q^x $$ and we might assume $y-1 = p^x$ and $y+1 = q^x$. But then $q^x-p^x=2$ but which is impossible for two different odd numbers $p,q,x \gt 1$ because $q^x-p^x = (q-p){ q^x-p^x\over q-p}$ and one factor must equal $1$ and the other must equal $2$.

  • Assume $y=2z+1$ being odd giving $ x^x = y^2-1 = 2z(2z+2)=8{z(z+1)\over 2}$. But the rhs being even means $x$ must be even. But this has already been handled above.

So none of the possible conditions on $x$ and $y$ allow a solution.

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