I was doing some basic Number Theory problems and came across this problem :
Find all primes that are the difference of the fourth powers of two integers
How can I go about it ?
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4$\begingroup$ Note that $x^4-y^4=(x^2-y^2)(x^2+y^2)$. $\endgroup$– André NicolasCommented Feb 18, 2015 at 17:45
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2$\begingroup$ Note that $x^2+y^2\ge 2,\forall x,y\in\mathbb Z$, so $x^4-y^4\in\mathbb P\iff x^2-y^2=1$ $\endgroup$– user26486Commented Feb 18, 2015 at 17:47
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1$\begingroup$ ... and $x^2=1+y^2$ has solution only $x=\pm 1, y=0$. $\endgroup$– vadim123Commented Feb 18, 2015 at 17:47
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1 Answer
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$x^4-y^4=(x^2-y^2)(x^2+y^2)$
Note that $x^2+y^2\ge 2,\forall x,y\in\mathbb Z$, so $$x^4-y^4\in\mathbb P\implies x^2-y^2=1\iff x-y=x+y=\pm 1\iff x=\pm 1, y=0$$
But $(\pm 1)^4-0^4=1\not\in\mathbb P$, so there are no such primes.
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$\begingroup$ Thanks a ton @user314 , for the proof by contradiction $\endgroup$– pranavCommented Feb 19, 2015 at 10:23