- State the binomial theorem.
- Prove that \((x+y)^5 \equiv x^5 + y^5 \pmod 5\).
- Prove by mathematical induction that \((x+y)^{5^n} \equiv x^{5^n} + y^{5^n} \pmod 5\) for every \(n \in \mathbb{N}\).
Solution
The binomial theorem states \[ (x+y)^m=\sum_{j=0}^m\binom mj x^{m-j}y^j. \] For \(m=5\), each intermediate binomial coefficient \(\binom5j\), \(1\le j\le4\), is divisible by \(5\). Hence \[ (x+y)^5\equiv x^5+y^5\pmod5. \]
We prove the final congruence by induction on \(n\). The case \(n=1\) is the preceding identity. If \[ (x+y)^{5^n}\equiv x^{5^n}+y^{5^n}\pmod5, \] then raising both sides to the fifth power preserves congruence, and applying the case \(m=5\) with \(X=x^{5^n}\) and \(Y=y^{5^n}\) gives \[ (x+y)^{5^{n+1}} \equiv (x^{5^n}+y^{5^n})^5 \equiv x^{5^{n+1}}+y^{5^{n+1}}\pmod5. \] Thus the claim holds for every \(n\ge1\).