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No Integer Solutions

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!

Note by Eeshan Upadhyay
3 years, 11 months ago

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Consider 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.

Aditya Parson - 3 years, 11 months ago

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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.

Mark Hennings - 3 years, 11 months ago

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Right, I forgot about that.

Aditya Parson - 3 years, 11 months ago

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@Aditya Parson Shouldn't you correct the solution?

Sagnik Saha - 2 years, 10 months ago

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