Prove that 3x^10 - y^10 = 1991 has no integer solutions.

I was hoping someone could give me a lead on how to approach this question.

Any help is much appreciated!

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TopNewestConsider the equation modulo \(11\).

Since \(11\) divides \(1991\) it must be that:

\[3x^{10}-y^{10} \equiv 0 \pmod{11}\]

By Fermat's Little Theorem: \(a^{10} \equiv 1\pmod{11}\), for integer \(a\).

Hence, \(3x^{10}-y^{10} \equiv 3-1 \equiv 2 \pmod{11}\).

Hence no solutions can exist.

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FLT only works for integers coprime to \(11\), so the other case you need to handle is the possibility that both \(x\) and \(y\) are multiples of \(11\). But this would imply that \(11^{10}\) divides \(1991\), which is not the case.

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Right, I forgot about that.

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