Freshman's Dream (Frobenius): Expanding pth Powers Modulo p

Explanations 4

• (x+1)pmod p≡xp+1(x+1)^p \mod{p} \equiv x^p+1(x+1)pmodp≡xp+1 Do a few numerical experiments. What is (x+1)2(x+1)^2(x+1)2? Hw about (x+1)3(x+1)^3(x+1)3? (x+1)5(x+1)^5(x+1)5? It's …

  By Jiazhen Tan, Victor Wang, Alex Federspiel

• Show that for all nnn, (x+1)pn≡xpn+1(modp)(x + 1)^{p^n} \equiv x^{p^n} + 1\pmod{p}(x+1)pn≡xpn+1(modp) Solution: It suffices to show that the coefficients of xkx^kxk for k∈(0,pn)…

  By Jonathan Xu, Insane Monkey, Moor Xu

• Taylor expansion derivation of the freshman's dream The freshman's dream is equivalent to the coefficient-wise congruence xp−1≡(x−1)p(modp),x^p - 1 \equiv (x-1)^p \pmod{p},xp−1≡…

  By Victor Wang

• Proof of the freshman's dream using xp−xx^p - xxp−x factorization modulo ppp From here we know that with coefficients interpreted modulo ppp, the polynomial xp−xx^p - xxp−x fact…

  By Victor Wang, Victor Wang