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I tried proving that $\dfrac{\Delta}{4}$ isn't a perfect square, but reducing it modulo 4 doesn't lead anywhere. Any help is greatly appreciated.

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    $\begingroup$ Hint: $\mod 5{}$ $\endgroup$
    – Wojowu
    Commented Sep 5, 2016 at 15:13
  • $\begingroup$ @Wojowu But there can be a situation where $x^4 + 131 \equiv 2 \pmod 5$ and $3y^4 \equiv 2 \pmod 5$ $\endgroup$
    – Airdish
    Commented Sep 5, 2016 at 15:15
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    $\begingroup$ Can you give an example? $\endgroup$
    – Wojowu
    Commented Sep 5, 2016 at 15:16
  • $\begingroup$ @Wojowu I honestly have no idea :P I just did the congruence calculation and arrived at the conclusion. $\endgroup$
    – Airdish
    Commented Sep 5, 2016 at 15:20
  • $\begingroup$ Redo your calculation. By Fermat's little theorem, $y^4\equiv 0$ or $1\pmod 5$, so $3y^4\equiv 0,3\pmod 5$. $\endgroup$
    – Wojowu
    Commented Sep 5, 2016 at 15:22

1 Answer 1

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$x^4\equiv0,1\pmod5$ for $x\in\Bbb Z$ and $131\equiv1\pmod5$. We now reduce the equation modulo 5 and check four cases, but all turn out to be impossible: $$0+1\not\equiv3\cdot0\pmod5$$ $$0+1\not\equiv3\cdot1\pmod5$$ $$1+1\not\equiv3\cdot0\pmod5$$ $$1+1\not\equiv3\cdot1\pmod5$$ $$x^4+131\not\equiv3y^4\pmod5$$ Hence the equation has no integer solution.

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