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How to prove or disprove following statement :

Conjecture :

Fermat number , $F_n=2^{2^n}+1$ , $(n \geq 2)$ is a prime number iff exists a unique representation of

$F_n$ in the form : $x^2+2\cdot y^2$ , where $\gcd(x,y)=1$ , $x,y \geq 0$ .

Assertion :

For every Fermat number $F_n$ , $(n \geq 2)$ it is true that : $F_n \equiv 1 \pmod 8$ .

Theorem :

Odd prime $p$ is expressible as : $p=x^2+2\cdot y^2$ iff

$p \equiv 1 \pmod 8$ , or $p \equiv 3 \pmod 8$ .


So , it follows that every Fermat prime $F_n$ , $(n \geq 2)$ is expressible as :$F_n=x^2+2\cdot y^2$

Question : How to prove uniqueness of this representation ?

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up vote 2 down vote accepted

The uniqueness follows because $R = \mathbb{Z}[\sqrt{-2}]$ is a Euclidean domain whose only units are $\pm 1.$ As you correctly observe, no rational prime $p$ congruent to $1$ or $3$ (mod 8) remains prime in $R.$ Then there is a prime $z$ in $R$ such that $z \overline{z} = p,$ and the only other primes $w \in R$ with this property are $-z, \overline{z}$ and $-\overline{z}.$ Hence if $z = a + b \sqrt{-2}$ for rational integers $a$ and $b,$ then $p = a^2 + 2b^2$ is the unique expression of $p$ in the form required with $a$ and $b$ positive. The conjecture is true. If $q$ is any prime which divides $F_{n}$ for $n >1,$ then we certainly have $q \equiv 1$ (mod $8$). If $F_n$ is divisible by more than one rational prime, it is a easy matter to combine the corresponding primes from $R$ and their complex conjugates in different ways to produce more than one representation of $F_n$ in the form $c^2 + 2d^2$ for rational integers $c$ and $d.$ If $F_n$ is a power higher than the first of a single rational prime, a similar argument can be applied.

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Although the conjecture is correct, it is not clear that it provides any further insight into checking whether $F_n$ is prime. –  Geoff Robinson Jan 5 '12 at 10:00
    
,I wrote primality test in Maple based on this conjecture but it doesn't seem to be practical because computation time is large... –  pedja Jan 5 '12 at 10:13
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Based on tests on first $5$ Fermat numbers, I found the following simple identity:

$F_n=2^{2^{n}}+1=(2^{2^{n-1}}-1)^2+ 2 \times (2^{2^{n-2}})^2$.

Thus, take $x=(2^{2^{n-1}}-1)$ and $y=(2^{2^{n-2}})$ and you have $F_n=x^2+2y^2$.

Then I found out this is already known as a recurrence relation satisfied by Fermat numbers.

http://en.wikipedia.org/wiki/Fermat_number

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So, in fact, the original question is equivalent to asking whether $(2^{2^{n-1}} -1) + 2^{2^{n-2}}\sqrt{-2}$ is a prime in the Euclidean ring $\mathbb{Z}[\sqrt{-2}]$ (see my answer). –  Geoff Robinson Jan 5 '12 at 11:05
    
yes, maybe. I say 'maybe' because I don't know abstract algebra. –  Nikhil Bellarykar Jan 5 '12 at 11:09
    
@GeoffRobinson I am thinking of a way to prove uniqueness but have failed so far. Can you please think of one that is 'elementary', in that it does not use group theory etc.? –  Nikhil Bellarykar Jan 5 '12 at 11:13
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I do not see that the original question of primeness or otherwise of $F_n$ is made any easier from this viewpoint. I see no obvious way of simplifying whether $F_n$ has another expression of the form $x^2 + 2y^2$ without checking all possiblities for $y$ up to the integer part of $\sqrt{\frac{F_n}{2}},$ except that $y$ must be even. –  Geoff Robinson Jan 5 '12 at 11:19
    
Primality or otherwise of $F_n$ is a different issue. I think that even if we prove the thing for all Fermat numbers,$2^{2^{n}}+1$, it should be ok. And, as you said, there ain't a simpler way to prove the thing anyway :) –  Nikhil Bellarykar Jan 5 '12 at 11:21
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