The most useful formula in competitions is the fact that \[a−b | a^n−b^n\] for all n, and
\[a+b | a^n+b^n\] for odd n.We have \[a^2−b^2=(a−b)(a+b)\].
But a sum of two squares such as \[x^2 + y^2\] can only be factored if **2xy** is also a
square. Here you must add and subtract **2xy**. The simplest example is the identity
of Sophie Germain:

\[a^4 + 4b^4 = a^4 + 4a^2.b^2 + 4b4 − 4a^2.b^2 = (a^2 + 2b^2)2 − (2ab)^2 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 − 2ab)\]

Some difficult Olympiad problems are based on this identity. For instance, in the 1978 Kurschak Competition, we find the following problem which few students solved.

example:1 \[n > 1 ⇒ n^4 + 4^n\] is never a prime. If n is even, then \[n^4 +4^n\] is even and larger than 2. Thus it is not a prime. So we need to show the assertion only for odd n. But for odd \[n = 2k + 1\], we can make the following transformation, getting Sophie Germain’s identity: \[n^4 + 4^n = n^4 + 4·4^{2k} = n^4 + 4 · (2k)^4\] which has the form \[a^4 + 4b^4\]. This problem first appeared in the Mathematics Magazine 1950. It was proposed by A. Makowski, a leader of the Polish IMO-team. Quite recently, the following problem was posed in a Russian Olympiad for 8th graders:

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TopNewestOk. I changed it.

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No you didn't. It is still the same.......

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Once check

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Did anyone understand this?

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Yeah I did........ although, there is a typo.....The question should be \(n^4+4^n\) instead of \(n^4+4n\)

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