Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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.

share|cite|improve this answer
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

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.

share|cite|improve this answer
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. – NikBels 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.? – NikBels Jan 5 '12 at 11:13
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 :) – NikBels Jan 5 '12 at 11:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.