In 1637 Fermat stated that $F_{5}=2^{2^{5}}+1=4294967297 $ is prime. On the contrary Euler showed that:
$F_{5}=(2^{16})^2+1^{2}=62264^{2}+20499^2$.
Because of Fermat I know that $F_{5}$ mod $4 $ is $1$. Does this help me a bit?
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1$\begingroup$ try the factor $$641$$ $\endgroup$– Dr. Sonnhard GraubnerNov 27, 2016 at 16:35
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$\begingroup$ math.stackexchange.com/questions/83938/… $\endgroup$– lab bhattacharjeeNov 27, 2016 at 17:30
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2 Answers
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Euler made this awesome computation :
We have $641 = 1 + 5 \times 2^7$. Therefore $5 \times 2^7 \equiv -1 \; [641]$. By squaring twice this congruence, we get : $5^4 \times 2^{28} \equiv 1 \; [641]$. However we also have $5^4 + 2^4= 641$. Therefore, $ - 2^4 \times 2^{28} \equiv 1 \; [641]$, which yields $ 1 + 2^{32} \equiv 0 \; [641]$. And since $32 = 2^5$, you are done showing that $F_5$ is not prime. Long live Euler's computations...
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$\begingroup$ This is an approach that does not use the fact of the squares. $\endgroup$ Nov 27, 2016 at 23:34
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Yes, of course the fact of $$F_5=\left(2^{16}\right)^2+1^2=62264^2+20449^2$$
Helps alot, only consider the following Theorem:
If a prime can be expressed as sum of two squares, then the representation is unique.
Then in your problem you have two different representations, then $F_5$ isn't a prime number.
You can see a proof of this theorem in the link in the comment.
answered Nov 27, 2016 at 21:15