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!

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!

No vote yet

5 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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. – Aditya Parson · 3 years, 8 months ago

Log in to reply

– Mark Hennings · 3 years, 8 months ago

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.Log in to reply

– Aditya Parson · 3 years, 8 months ago

Right, I forgot about that.Log in to reply

– Sagnik Saha · 2 years, 8 months ago

Shouldn't you correct the solution?Log in to reply