Sum of squares? That sounds familiar

Number Theory Level 3

I've just found the newest largest known prime number $2^{p}-1$ . Can it be expressed as the sum of 2 perfect squares $a^{2}+b^{2}$ ?

Nondeterministic; doesn't depend on the value of

p

Deterministic; depends on the value of

p

No Yes!

2 solutions

Abdullah Shahriar
Jan 3, 2015

$2^p-1 \equiv 3 \pmod{4}$

$a^2 \equiv 0,1 \pmod{4}$

$b^2 \equiv 0,1 \pmod{4}$

$a^2+b^2 \equiv 0,1,2 \pmod{4}$

So $2^p -1$ can not be expressed as $a^2+b^2$ $\Rightarrow \boxed{Proved}$

Yeah, this was my other proof. Also , it is necessary that $a^{2}$ be odd and $b^{2}$ be even to preserve parity, so it can never be $1 \pm 1 \pmod{4}$ .

Jake Lai - 6 years, 5 months ago

Jake Lai
Jan 3, 2015

Fermat's two-square theorem states that a prime can be expressed as the sum of two squares iff it is of the form $4n+1$ , and hence my prime $4n+3 = 4(2^{p-2}-1)+3$ cannot be expressed as the sum of two squares.

Can we make conclusions from this?

$\dfrac{2^p - 1}{2 - 1} = 1 + 2 + 2^2 + 2^3 +.... 2^{p-1}?$

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U Z - 6 years, 5 months ago

Ah, sorry about the link problem.

To answer your question, yes? But the only way I could fathom utilising that is noting that $4 | 2^{n}$ for $n \geq 1$ and the rest of the terms add to 3, which makes the conversion to a geometric partial sum trivial.

Jake Lai - 6 years, 5 months ago

Sum of squares? That sounds familiar

2 solutions

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