Imo 1964-1

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Masum
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Imo 1964-1

Unread post by Masum » Thu Dec 09, 2010 8:53 pm

Prove that $n^4+4^n$ is not a prime for $n>1$
One one thing is neutral in the universe, that is $0$.

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Moon
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Re: Imo 1964-1

Unread post by Moon » Sun Jan 09, 2011 10:28 pm

This is really a nice problem!
Consider two cases:
1. $n$ even. Sophie is inviting you!
2. $n$ odd. My old good old friend FLT.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

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nafistiham
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Re: Imo 1964-1

Unread post by nafistiham » Tue Apr 03, 2012 9:20 am

for $n=2k$, $n^4+4^n$ is divisible by $2$
for $n=2k+1$

\[\begin{align*}
n^4+4^n &=(2k+1)^4+4^{2k+1} \\
&=(2k+1)^4+4\cdot4^{2k} \\
&=(2k+1)^4+4\cdot2^{4k} \\
&=(2k+1)^4+4\cdot(2^k)^4
\end{align*}\]

sophie germain
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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