I am interested in the possible natural solutions of the equation $a^n + b^n = c^2$ where $n \geq 4$ is fixed. I am not sure if it is well-known or not, so any suggestion would be helpful.
3 Answers
To complement Boris Novikov's answer, Darmon and Merel have shown that the equation has no solution with $\gcd(a,b,c)=1$. See J. reine angew. Math. 490 (1997), 81-100.
$2^{2n+1}+2^{2n+1}=( 2^{n+1})^2$.
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$\begingroup$ a solution when the exponent in the problem is odd. $\endgroup$– rinoDec 11, 2013 at 19:12
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3$\begingroup$ Sorry, I don't understand your comment. $\endgroup$ Dec 11, 2013 at 20:18
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1$\begingroup$ More generally, choose arbitrary $p$ and $q$. Set $r=p^{2n+1}+q^{2n+1}$. Then $a=pr$, $b=qr$ and $c=r^{n+1}$ gives a solution for exponent $2n+1$. $\endgroup$– user25199Dec 11, 2013 at 21:47
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2$\begingroup$ @Boris: rino meant that in your example the exponents on the left hand side are odd. As Carl mentioned, there is no example with even exponents. $\endgroup$ Dec 11, 2013 at 23:36
Assuming $gcd(a,b,c)=1$ there are no solutions (see GH's answer), and this fits well for the hyperbolic case of the generalised Fermat equation $x^p+y^q=z^r$ with $1/p+1/q+1/r<1$, and $gcd(x,y,z)=1$, because for $n>4$ above we are in this case. By Darmon-Granville theorem we know that there are only finitely many solutions, and conjecturally at most $10$ solutions. None of these is for $(p,q,r)=(n,n,2)$ with $n>4$.