Prove that $x^2 + y^2 = 3(z^2 + m^2)$ has no solutions in integer [closed]

Solution 1:

As 3 divides the right hand side, one must have $$x^2+y^2=0 \mod 3.$$ But $$0^2=0\mod 3,$$ $$1^2=1\mod 3,$$ $$2^2=1\mod 3,$$ so the only way to have $x^2+y^2=0\mod 3$ is if $x=y=0\mod 3$. We can then write $x=3x_1$ and $y=3y_1$, so $$3(x^2_1+y^2_1)=z^2+m^2.$$ But this is just the original problem with new variables, so we get $z=3z_1$ and $m=3m_1$. Unless $x=y=z=m=0$, this process will continue indefinitely, which is impossible, as 3 can't divide any positive integer infinitely many times.

Conclusion: $x=y=z=m=0$ is the only integer solution.

Solution 2:

Instead of modulo $3$, one can work modulo $8$: $$a^2\equiv 0,1,4\pmod{8}.$$ That means $$a^2+b^2\equiv 0,1,2,4,5\pmod{8}.\tag{1}$$ From (1) $$3(z^2+m^2)\equiv 0,3,6,4,7\pmod{8}.\tag{2}$$ Compare (2) and (1) we see that $$x^2+y^2=3(z^2+m^2)\equiv 0,4\pmod{8}$$ which only happens when all $x,y,z,m$ are even. We then can continue by reduction.

Solution 3:

Hint: assume that $(x,y,z,m)$ is a minimal solution. Then prove that $$ 3|(x^2 + y^2)\implies 3|x\ \ \ \& \ \ \ 3|y $$