While trying to get around this question, which is the positive integers solutions to $x^3=y^5+100$, I did some simple manipulations to get: $$q|y\implies x^3\equiv100\pmod q\\ p|x \implies y^5\equiv-100 \pmod p$$ Then I noticed that there is not always solutions to that equations. For example, mod 7 we have $0^3\equiv0, 1^3\equiv 1,2^3 \equiv 1, 3^4\equiv 6, 4^3\equiv 1, 5^3\equiv 6, 6^3 \equiv 6$ but $100 \equiv 2$.

My question is: Is there a way to characterize the numbers $q$ that the equation has no solutions $\mod q$?

What I did so far is brute-forcing the solutions $q\not= \{7, 8, 13, 19, 31, 43, 61, 67, 97, 109, 125, 151, 157, 163, 181, 193, 199, 211, 223, 229, 241, 277, 283, 307, 313, 337, 367, 373, 379, 397, 409, 433, 439, 487, 499, 523, 541, 571, 577, 601, 619, 631, 709, 727, 757, 769, 787, 811, 823, 853, 877, 883, 919, 937, 991\}$

And $p\not= \{8, 31, 41, 61, 71, 125, 131, 151, 181, 191, 211, 241, 271, 311, 331, 401, 421, 431, 461, 491, 541, 571, 601, 631, 661, 691, 701, 751, 761, 811, 821, 881, 911, 941, 971, 991\}$


By the Chinese Remainder Theorem, if $m={q_1}^{a_1}\cdots{q_r}^{a_r}$, then the congruence is solvable if and only if it is solvable modulo the prime powers dividing $m$. Therefore it's enough to consider prime powers as moduli.

Let's start off with primes. If a prime is $2$ modulo $3$, then all numbers are cubic residues. To see why, $a^3\equiv b^3$ implies $(a^3)^{\frac{q+1}3}\equiv(b^3)^{\frac{q+1}3}$, which is $a^2\equiv b^2$ by Fermat's little theorem. Thus either $a\equiv b$ or $a\equiv-b$, but the latter is impossible unless $a\equiv b\equiv0$. Hence all $q$ cubic residues are different.

Thus if $q\equiv2\pmod3$ then $100$ is a cubic residue modulo $q$. When extending this to prime powers, the only problem is when $5^3\mid m$. Any cube that is divisible by $5$ is congruent to $0$ modulo $125$, but certainly not congruent to $100$.

The same problem appears when $q=2$. If $2^3\nmid m$ there is no problem yet, but if $2^3\mid m$ the congruence is impossible for the same reason.

If $q\equiv1\pmod3$ the answer isn't simple at all. Unlike quadratic residues, there aren't yet beautiful laws like Gauss' Lemma or Quadratic Reciprocity that permit us to easily compute 'cubic Legendre Symbols'. Very little is known about cubic residues modulo primes $1$ modulo $3$. You may want to check out this Wikipedia page, but you won't find a complete answer there. Especially not because $100$ is not a prime. There are a few theorems about the cubic character of $2$ (perhaps of $5$ too) but I'm not sure if there's a multiplicative property as there is with quadratic residues. And even if there were, with the theorems provided so far the final answer would be too complicated for practical use.

I think I should not elaborate on quintic residues (or whatever we call them). As is the case $q\equiv-1\pmod3$, here every integer will be a quintic residue if $p\equiv-1\pmod5$. The cases where $p\not\equiv-1\pmod5$ are hardly investigated so far.

  • $\begingroup$ So sad story! Especially when I discovered independently the 2 and 5 does not divide x and y )-:. $\endgroup$ Feb 9 '14 at 12:34
  • $\begingroup$ I find it sad too. But nonetheless it motivates me to investigate this myself, as soon as I feel experienced enough for that. Perhaps I first should pass a few more years at university ;-) $\endgroup$ Feb 9 '14 at 12:39

It has a solution $x=7, y=3$,

If $p-1$ is not divisible by 3 or 5, and is modulo 2 mod 4, this always has a unique solution, usually the example given above. This means because there are an infinity of such primes, which makes this unique, it means that this solution is unique.


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