The Proofathon Spring Competition was a huge success thanks to you guys! This time we had two perfect scorers and the largest turn out we've ever had! The April Monthly: Combinatorics is already underway, so visit the Proofathon Contests Page for your chance to compete. If you are new to Proofathon, sign up at Proofathon.org and be sure to check out our Facebook for updates and news. Here's a problem from our last competition:

Find all polynomials \(p(x)\) such that for all real numbers \(a,b,c \neq 0\) satisfying \(\frac{1}{a}+\frac{1}{b} = \frac{1}{c}\), \[\frac{1}{p(a)}+\frac{1}{p(b)} = \frac{1}{p(c)}\]

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## Comments

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TopNewestWhen i took the contest, my solution was plug in a=b, then rearrange to 2f(x)=f(2x), implying that f(x) is linear and so p = ax. Then plug in and show it works. Maybe my proof was too simple and had a flaw though....

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Okay... I'm not sure whether this is super easy or really tough. I'm going to assume it's easy.

We can prove that for any class of polynomials \(p_n(x)=\frac{x}{n}\) that the conditions will be satisfied. Multiplying \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) by \(n\), we achieve \(\frac{1}{\frac{1}{n}a}+\frac{1}{\frac{1}{n}b}+\frac{1}{\frac{1}{n}c}\), which is in the form given as long as \(p(x)=\frac{x}{n}\) which we've already stated. That's all I can think of. There's probably something I'm missing, though. @mention me if you have a cool proof! :D

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