How can I prove this? For every integer m with 2 divides m and 4 does not divide m, there are no integers x and y that satisfy x^2+3y^2=m.

Question

I know that m is even and m/2 is odd, but I don't know where/how I can use this. Also, 3y^2 is odd and the sum is odd when x^2 is even. I'm trying to prove that its always odd, but I'm stuck. Can someone please help? Thanks

Please don't have part of your question only in the title – Natecat Dec 11 '16 at 23:34 — Natecat, Dec 11 '16 at 23:34

score 1 · Accepted Answer · answered Dec 11 '16 at 23:47

1

Suppose : $m=x^2+3y^2$ is even.

Case $1$ : $x$ is even. Then $y$ must be even as well. So, $m$ is divisble by $4$.

Case $2$ : $x$ is odd. Then, $\ x^2\equiv 1 \mod 4\ $. Since $y$ must be odd as well, we also have $\ y^2\equiv 1\mod 4\ $. Hence, $x^2+3y^2\equiv 0\mod 4$.

So, again, $m$ is divisble by $4$

answered Dec 11 '16 at 23:47

Peter

84,454

It is given that m is not divisible by 4. The question is asking to prove that the equation x^2 + 3y^2 = m has no solution. – User2345 Dec 11 '16 at 23:50
1

I have done a contradiction-proof. If $m$ is divisble by $2$ and of the form $x^2+3y^2$, then it must be divisble by $4$. Hence, if $n$ is divisble by $2$, but not by $4$, it cannot be of the form $x^2+3y^2$ – Peter Dec 11 '16 at 23:51

score 0 · Answer 2 · answered Dec 12 '16 at 00:10

0

Hint $\ {\rm mod}\,\ 2\!:\ x\equiv y\ \Rightarrow\ {\rm mod}\,\ 4\!:\ x^2\equiv y^2\equiv -3y^2$

answered Dec 12 '16 at 00:10

Bill Dubuque

272,048

How can I prove this? For every integer m with 2 divides m and 4 does not divide m, there are no integers x and y that satisfy x^2+3y^2=m.

2 Answers2