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!

#HelpMe! #Advice #MathProblem #Opinions

Note by Eeshan Upadhyay
7 years, 6 months ago

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5 votes

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Comments

Consider the equation modulo 1111.

Since 1111 divides 19911991 it must be that:

3x10y100(mod11)3x^{10}-y^{10} \equiv 0 \pmod{11}

By Fermat's Little Theorem: a101(mod11)a^{10} \equiv 1\pmod{11}, for integer aa.

Hence, 3x10y10312(mod11)3x^{10}-y^{10} \equiv 3-1 \equiv 2 \pmod{11}.

Hence no solutions can exist.

Aditya Parson - 7 years, 6 months ago

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FLT only works for integers coprime to 1111, so the other case you need to handle is the possibility that both xx and yy are multiples of 1111. But this would imply that 111011^{10} divides 19911991, which is not the case.

Mark Hennings - 7 years, 6 months ago

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Right, I forgot about that.

Aditya Parson - 7 years, 6 months ago

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@Aditya Parson Shouldn't you correct the solution?

Sagnik Saha - 6 years, 6 months ago
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