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Residue Classes

Consider the diophantine equation -

\( x^3 + y^4 = 7 \)

The solution which was given in the book had argument starting like - "Consider the residue class of \(x^3\) modulo 13"

From where does one get motivation to check the residue class of that particular modulo?

Easy ones can be seen directly like checking residue class of modulo 3 in case of squares, modulo 7 in case of cubes, but what about others?

And yes, is there any list available of frequently used residue classes of modulo x? I think that might help many students. :)

Note by Soham Chanda
3 years, 8 months ago

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You are looking for prime numbers that have a lot of cube roots and fourth roots of unity. Modulo \(13\), there are \(3\) cube roots of unity (\(1,3,9\)) and \(4\) fourth roots of unity (\(1,5,8,12\)), and so there are exactly \(4\) cubes and \(3\) fourth powers modulo \(13\). This cuts down the number of possible residues of \(x^3+y^4\) down to size (and, specifically, misses \(7\), even if that is the only residue that gets missed out). Mark Hennings · 3 years, 8 months ago

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@Mark Hennings Great explanation! To expand slightly, the primitive element theorem tells us that modulo \(p\) there is a multiplicative generator \(g\), such that the order of \(g\) is \(p-1 \). Hence, we would have 3 cube roots if and only if \(p-1 \) is a multiple of 3. (Otherwise, there is only 1 cube root, namely 1.) In turn, this implies that there are \( \frac{p-1}{3} \) possible values of cubes (and otherwise, there are \(p-1\) values).

And if 13 didn't work, what would be the next number we try, given the same motivation? Calvin Lin Staff · 3 years, 8 months ago

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@Calvin Lin \(37\), being the next prime of the form \(12n+1\). Mark Hennings · 3 years, 8 months ago

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I think for cubes always check modulo 7,9, and 13 because the cubes leave very less numbers of distinct remainders modulo these numbers. Kishan K · 3 years, 8 months ago

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