×

# 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!

3 years, 8 months ago

Sort by:

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. · 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. · 3 years, 8 months ago

Right, I forgot about that. · 3 years, 8 months ago