How to prove there are no solutions to the equation $a^2-4ab+b^2=0$ if $a$ and $b$ are real numbers and $b \neq 0$? I am trying to prove $a^2-4ab+b^2=0$ has no solutions for all real numbers $a$ and $b$ and $b \neq 0$
My attempt:
We know that $a^2 \geq 0$ and $b^2 > 0$ since $b \neq 0$. So then $a^2 + b^2 >0$. Now I'm stuck as I'm not sure how to show that $a^2 + b^2 = 4ab$ has no solutions given the above conditions. A little assistance would be greatly appreciated.

This isn't true. Consider $a = 1,$ and $b = 2 + \sqrt{3}$. 
Then $a^{2} -4ab + b^{2} = 1 -4(2 + \sqrt{3}) + (2 + \sqrt{3})^{2} = 1 - 8 -4\sqrt{3} + 4 +4\sqrt{3} +3 = 0$.
Perhaps the question meant to ask over the integers instead of the reals?
