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Iranian Mathematical Olympiad

Number theory Difficulty 5.8 AIME, harder Prove it Iran

p≠13p \neq 13 is a prime number in the form of 8k+58k + 5 for some natural number kk, and 3939 is a non-residue modulo pp. Prove that equation x14+x24+x34+x44≡0(modp)x_1^4 + x_2^4 + x_3^4 + x_4^4 \equiv 0 \pmod{p} has a solution in the set of integers such that p∤x1x2x3x4p \nmid x_1x_2x_3x_4.

Solution

Note that
∑1≤xi≤p−1(1−(x14+x24+x34+x44)p−1)=p(p−1)4−(p−1)4(∑p−14∣θi;∑θi=p−1A(p−1θ1,θ2,θ3,θ4))⏞p. \sum_{1 \le x_i \le p-1} \left(1 - \left(x_1^4 + x_2^4 + x_3^4 + x_4^4\right)^{p-1}\right) \stackrel{p}{=} \qquad (p-1)^4 - (p-1)^4 \overbrace{\left(\sum_{\substack{\frac{p-1}{4}|\theta_i; \\ \sum \theta_i = p-1}}^A \begin{pmatrix} p-1 \\ \theta_1, \theta_2, \theta_3, \theta_4 \end{pmatrix}\right)}^{p}.

A=p4(p−1p−1)+12(p−1p−14,3p−14)+12(p−1p−14,p−14,p−12)+(p−1p−14,p−14,p−14,p−14)=p4+12(p−1)!(p−14)!(3(p−1)4)!+6(p−1)!(p−12)!2+12(p−1)!(p−12)!(p−14)!2+(p−1)!(p−14)!4=p4+12(p−1)!(−1)p−14(p−1)!+6(p−1)!−1+12(p−1)!(p−12)!(p−14)!2+(p−1)!(p−14)!4=p4−12+6+12(p−12)!(p−14)!2+−1(p−14)!2.\begin{align*} A \stackrel{p}{=} & 4 \binom{p-1}{p-1} + 12 \binom{p-1}{\frac{p-1}{4}, 3\frac{p-1}{4}} + 12 \binom{p-1}{\frac{p-1}{4}, \frac{p-1}{4}, \frac{p-1}{2}} + \binom{p-1}{\frac{p-1}{4}, \frac{p-1}{4}, \frac{p-1}{4}, \frac{p-1}{4}} \\ \stackrel{p}{=} & 4 + \frac{12(p-1)!}{\left(\frac{p-1}{4}\right)!\left(\frac{3(p-1)}{4}\right)!} + \frac{6(p-1)!}{\left(\frac{p-1}{2}\right)!^2} + \frac{12(p-1)!}{\left(\frac{p-1}{2}\right)!(\frac{p-1}{4})!^2} + \frac{(p-1)!}{\left(\frac{p-1}{4}\right)!^4} \\ \stackrel{p}{=} & 4 + \frac{12(p-1)!}{(-1)^{\frac{p-1}{4}}(p-1)!} + \frac{6(p-1)!}{-1} + \frac{12(p-1)!}{\left(\frac{p-1}{2}\right)!(\frac{p-1}{4})!^2} + \frac{(p-1)!}{\left(\frac{p-1}{4}\right)!^4} \\ \stackrel{p}{=} & 4 - 12 + 6 + 12 \frac{\left(\frac{p-1}{2}\right)!}{\left(\frac{p-1}{4}\right)!^2} + \frac{-1}{\left(\frac{p-1}{4}\right)!^2}. \end{align*}
If$a≡p(p−12)!$and$b≡p1(p−14)!2$,then$A≡p−2+12ab−b2$;therefore, If \$a \stackrel{p}{\equiv} (\frac{p-1}{2})!\$ and \$b \stackrel{p}{\equiv} \frac{1}{(\frac{p-1}{4})!^2}\$, then \$A \stackrel{p}{\equiv} -2 + 12ab - b^2\$; therefore,
-A + 1 ≡p\stackrel{p}{\equiv} b^2 - 12ab + 3 ≡p\stackrel{p}{\equiv} (b - 6a)^2 - 36a^2 + 3 ≡p\stackrel{p}{\equiv} (b - 6a)^2 + 39.

And according to the assumption, (−39p)=−1\left(\frac{-39}{p}\right) = -1, which completes the proof.

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