#Number Theory

For discussing Olympiad Level Number Theory problems
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#Number Theory

Unread post by Kazi_Zareer » Sat Sep 26, 2015 3:38 pm

If the square of any prime number can be expressed as $a^2 + 2 b^2$, then show that the prime number can be written as $x^2 + 2 y^2$.
Last edited by Masum on Sat Oct 10, 2015 12:21 am, edited 1 time in total.
Reason: Put dollars between math expressions

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Re: #Number Theory

Unread post by Masum » Sat Oct 10, 2015 12:23 am

See the article here.
One one thing is neutral in the universe, that is $0$.

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Re: #Number Theory

Unread post by nayel » Sun Nov 22, 2015 7:29 am

Your claim does not hold if $b=0$ so I'll assume that $ab\neq 0$. Then $p$ must be odd.

Suppose that $p^2=a^2+2b^2$. Then $(p+a)(p-a)=2b^2$, so $p\pm a$ are both even. Set $p+a=2x$ and $p-a=2y$. Then $2xy=b^2$, implying that $b$ is even. Set $b=2z$. Then $xy=2z^2$.

Note that $(x,y)$ divides $2p$ and $2a$, hence $2(p,a)$. Since $p$ cannot divide $a$, $(p,a)=1$, i.e., $(x,y)=1$ or $2$. If $(x,y)=2$ then $4$ divides $p\pm a$, and so $4$ divides $2p$, which is impossible.

Hence $(x,y)=1$. Since $2\mid xy$, it follows that $2\mid x$ or $2\mid y$. If $x=2x'$ then $x'y=z^2$ with $(x',y)=1$. Hence $x'=u^2$, $y=v^2$ for some integers $u,v$. This implies $x=2u^2$, $y=v^2$ and thus $p+a=4u^2$, $p-a=2v^2$. Solving for $p$ yields $p=v^2+2u^2$, as desired. If $y=2y'$ one obtains a similar conclusion.
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

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