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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.

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    $\begingroup$ Try looking for counterexamples. You can't prove a false statement. $\endgroup$ – David K Dec 12 '16 at 17:53
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    $\begingroup$ Maybe you're looking for a lack of rational/integer solutions? Then solving the quadratic (basically) does the trick... $\endgroup$ – Steven Stadnicki Dec 12 '16 at 17:57
  • $\begingroup$ but it has integer solutions for some real values of $a$ and $b$. $\endgroup$ – Jorge Dec 12 '16 at 17:59
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This has real solutions.

Solve the quadratic equation for $a$ and you find

$a = (2 \pm \sqrt{3}) b$

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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?

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since $$b\ne 0$$ we can divide by $b$ and we get $$\frac{a^2}{b^2}-4\frac{a}{b}+1=0$$ Setting $$\frac{a}{b}=t$$ you will got a quadratic equation. Can you proceed?

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Complete the square and write it as:

$$ (a-2b)^2 = 3 b^2 \\ a-2b = \pm \sqrt{3}\,b \\ a = (2 \pm \sqrt{3})b $$

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See if you can find all integer solutions to $$ a^2 - 4 a b + b^2 = 1. $$ You might as well stick with $a,b \geq 0.$ If we had $a \geq 1, b \leq -1,$ or $a \leq -1, b \geq 1,$ we would then have $a^2 - 4ab + b^2 \geq 6.$ So, if both nonzero, either both positive or both negative.

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notice that $a^2-4ab+b^2$ is a polynomial of degree $2$ with variable $a$.

So we can solve it with the quadratic formula.

You get $\frac{4b\pm\sqrt{16b^2-4b^2}}{2}=(2 \pm \sqrt{3}) b$, which in fact is always a real number.

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  • $\begingroup$ oh yeah, my bad, thanks. $\endgroup$ – Jorge Dec 12 '16 at 18:00

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