Fermat's Little Theorem is the habit of reducing powers modulo primes using $a^{p-1}\equiv 1$ when $a$ is not divisible by $p$. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
Prove that for every prime $p$, the number $p$ divides $\binom{2p}{p} - 2$.
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By Fermat's little theorem in the form $a^p \equiv a \pmod p$ for all integers $a$, the binomial $(1 + x)^p \equiv 1 + x^p \pmod p$ as polynomials with integer coefficients (the freshman's dream, since $p \mid \binom{p}{k}$ for $0 < k < p$).
Generalization: Fermat is the prime case of Euler's theorem $a^{\varphi(n)} \equiv 1 \pmod n$ for $\gcd(a,n) = 1$ (with $\varphi(p) = p-1$). The cleaner sibling $a^p \equiv a \pmod p$ holds for ALL integers $a$, including multiples of $p$, and generalizes to $a^{p^k} \equiv a^{p^{k-1}} \pmod{p^k}$.
Where it appears: collapsing enormous exponents in $a^N \bmod p$, power-tower problems (reduce the tower from the top using the exponent's modulus), primality testing (Fermat / Miller–Rabin), and polynomial-coefficient arguments via $(1+x)^p \equiv 1 + x^p$.
Pitfall: Fermat requires $\gcd(a, p) = 1$ for the $a^{p-1} \equiv 1$ form — if $p \mid a$ then $a^{p-1} \equiv 0$, not $1$. When reducing the exponent of $a^E \bmod p$, reduce $E$ modulo $p - 1$ (or the order of $a$), NOT modulo $p$; and the converse fails — $a^{n-1}\equiv 1 \pmod n$ does not prove $n$ prime (Carmichael numbers like $561$ fool it).
Spot the Signal
Look for problems where the key step is reducing powers modulo primes using $a^{p-1}\equiv 1$ when $a$ is not divisible by $p$.
You can describe the hard part as reducing powers modulo primes using $a^{p-1}\equiv 1$ when $a$ is not divisible by $p$, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the integer constraint that calls for fermat's little theorem, then rewrite the givens around it.
Name the relation that makes Fermat's Little Theorem legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Fermat's Little Theorem just because the surface looks familiar; verify the required condition first.
Applying Fermat's Little Theorem because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let fermat's little theorem reveal the integer structure; then compute only what remains.
Try it on:
Practice a contest problem where the key step is reducing powers modulo primes using $a^{p-1}\equiv 1$ when $a$ is not divisible by $p$.