I am used to solving a lot of Mathematical Olympiad problems, but the most rememberable problem that I love is a numerical problem. It took me three hours to solve it. Why is it beautiful.? Because it is solved by some wonderful techniques:

• Reduce to absurdity
• Euler’s theorem
• Lifing expoinment lemma
• Some simple numerical inferences

Problem

Consider that \(a,b,c\) are natural numbers, \(a,b\) are different from \(c\). Show that there exists infinitely prime number \(p\) such that exists an integer \(n\) to satisfy \(p | a^n+b^n-c^n\)

Proof

Consider that \((a,b,c)=1\) because if \((a, b, c) = q\), we have a new sub-problem \((a', b', c')\) where \(a' = \frac{a}{q}, b' = \frac{b}{q}, c' = \frac{c}{q}\).

Assuming that the set the prime divisors of \(a^n+b^n-c^n\) is limited and it is difined as \(X=\{p_1,p_2,...,p_k \}\).

Assuming that \((a,p_i)=(b,p_i)=(c,p_i)=1 \ \forall{1 \le i \le k}\). Choosing \(n= \phi(p_1)\phi(p_2)....\phi(p_k)\). Then apply Euler’s theorem, we have:

\[a^n \equiv 1 \pmod {p_i}\] \[b^n \equiv 1 \pmod {p_i}\] \[c^n \equiv 1 \pmod {p_i}\]

Hence \(a^n+b^n-c^n \equiv 1 \pmod {p_i}\). Contradictory!

So that exist some numbers \(i \in (1,2,....,k)\) that \(p_i | a\) or \(p_i | b\) or \(p_i | c\)

We define: \(A=\{i ; p_i |a \}\)

\[B=\{i ;p_i |b \}\] \[C=\{i ; p_i |c \}\]

Apply Lifing expoinment lemma, we can choose \(n_1\) big enough so that:

\[v_{p_i}(b^{n_1}-c^{n_1}) < v_{p_i}(a^{n_1}) (i \in A)\] \[\Rightarrow a^{n_1}+b^{n_1}-c^{n_1}= \prod_{i \in A}p_i^{v_{p_i}(b^{n_1}+c^{n_1})}(P_1-Q_1)\]

where \(p_i | P_1 ; p_i \not| Q_1\) for \(i \in A\)

Similar as the thing above, exist \(n_2\) so that:

\[a^{n_2}+b^{n_2}-c^{n_2}= \prod_{i \in B}p_i^{v_{p_i}(a^{n_2}-c^{n_2})}(P_2-Q_2)\]

where \(p_i | P_2;p_i \not| Q_2\) for \(i \in B\)

And exist \(n_3\) so that:

\[a^{n_3}+b^{n_3}-c^{n_3}= \prod_{i \in C}p_i^{v_{p_i}(a^{n_3}+b^{n_3})}(P_3-Q_3)\]

where \(p_i | P_3;p_i \not| Q_3\) for \(i \in C\)

Finally, choose:

\(n=n_1n_2n_3\prod\phi(p_i)\) for \(i \not\in A,B,C\)

We have \(a^n+b^n-c^n\) could be written:

\[\prod{p_i}^{\alpha_i}(P-Q)\]

where \(\alpha_i \ge 1 (i \in A,B,C)\)and \(p_i \not| P-Q (i=1,2,...,k)\)

\(\Rightarrow\) Exist a prime number \(q \not\in X\) that satisfies \(q|P-Q|a^n+b^n-c^n\).

It is a contradiction. The problem is solved.