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I'm a beginner and I've only recently picked up "congruence relations". I recently asked a question about intuition for congruence relations, but I figured that maybe if I just do some problems, I'll see why reducing a number to their remainder is helpful sometimes.

I came across a particularly interesting question: Find integer solutions, if any, to $a^4 - 3b^2 =11$ and honestly at first glance, I didn't think it had anything to do with congruence relations but boy was I wrong.

For any integers, $a$ and $b$, $a^4 \equiv 0 \text{ or } 1 \, (\bmod 3)$ and $-3b^2 \equiv 0 \, (\bmod 3)$. Hence $a^4 -3b^2 \equiv 0 \text{ or } 1\, (\bmod 3)$ but $11 \bmod 3= 2$ so there are no integer solutions.

Elegant. But wait, I understand why $3$ is used. But how does one think of $3$ and know that $3$ would work?

In certain situations, picking a modulus is easy. For instance, you want to know the last $k$- digits of a number (in decimal system), cool just do $\bmod 10^k$. You want to show that the cube of any integer is either of the form $9k, 9k+1$ or $9k+8$, choose $\bmod 9$.

My Question: What should be my thought process or what should I look for while deciding a modulus? From your experience, how do you figure out what you want to "wrap" your numbers around with?

William
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  • "Deciding the modulus" is not always obvious, but here it is clear that modulo $3$ the equation is only $a^4=11=2$, which has no solution modulo $3$. – Dietrich Burde Aug 21 '21 at 11:57
  • @DietrichBurde Yes, you're right. But I fully understand it's not always obvious. My question (it's in the end) is not that I want to find a magical way to figure out the modulus, no that'd be unreasonable frankly. I understand it comes with experience, which is why I ask the "thought process" of an experienced person. I'm really really really a beginner, in that I've literally no experience working with congruence relations. – William Aug 21 '21 at 12:04
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    I guess the thought process here would be to remove terms to simplify the equation. Consider the expression $ab + cd = e$. Taking this modulus $c$ or $d$ you can remove the $cd$ term entirely. Similarly, taking this modulus $a$ or $b$ you can remove the $ab$ term entirely. Since $-3$ is the coefficient of $b^2$ it is only natural to try $\mod 3$. Equations of the forms $a^n = b \mod c$ are common and well-studied so trying to get it in this form is always helpful. – egglog Aug 21 '21 at 12:10
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    There is no guarantee a Diophantine polynomial equation can be solved using congruence, but it probably as a first try to get the "feel" of the problem is to try some small $n$, or $n$ related to the powers (e.g. all powers are 0 or 1 mod $\phi(n)$), or prime $p$ that make some/many coefficients disappear. – user10354138 Aug 21 '21 at 12:13

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Often in Diophantine equations containing $n$th powers it is useful to pick a modulus such that the $n$th powers don't yield many values. The way these moduli are picked is by utilising Fermat's Little Theorem/Euler's Totient Theorem, which guarantees the power will generate a very small subset of possible remainders in your chosen modulus.

A useful rule of thumb I follow is if the power is $n$, then use a modulus $m$ such that $\phi(m) = n$ or $2n$ - Euler's Theorem guarantees the remainder is in $\{0, 1\}$ or $\{0, \pm 1\}$ respectively.

With your question for example, the power $4$ and the power $2$ tells me to use $\pmod{3}$, and the fact it is $3b^2$ reinforces this idea.

Sharky Kesa
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