# Sophie Germain’s identity(divisibility,number theory)

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: Note by Chakravarthy B
10 months, 3 weeks ago

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Did anyone understand this?

- 10 months, 2 weeks ago

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

- 10 months, 2 weeks ago

Ok. I changed it.

- 10 months, 2 weeks ago

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

- 10 months, 2 weeks ago

Once check

- 10 months, 2 weeks ago

Yup, now it is fine....!!

- 10 months, 2 weeks ago