Solve $b^2 = 8b - 16a$ for $a,b\in\Bbb Z$

Question

I want to know when the equation : $$b^2 = 8b - 16a$$ is true.

I wanted to create a polynom like this one : $$(b-4)^2 = 16(-a+1)$$ but then I don't know what to do. Any idea?

Answer 1

The equation is actually $$(b-4)^2=16(1-a)$$

Your LHS is a square number, your RHS is $16 x$ where $x$ is an integer. When can your RHS be made into a square number?

It already has $2^4=(2^2)^2$ as a factor, so all you need is for $x$ to be a square number.

So $x=1-a=n^2$ for $n\in \Bbb N$. Then $b-4=\pm4n$.

So solutions are $$a=1-n^2\\b=4\pm4n$$ $\forall n\in \Bbb N$.

Answer 2

we get $$(b-4)^2=16(1-a)$$ i hope this solves your problem

Answer 3

You could just use the common second degree equation formula.

So:

$$ b^2-8b+16a=0\implies b = \frac{8 \pm \sqrt{64-64a}}{2}$$

From this we can conclude that there will only exist a solution over $\mathbb{R}$ when $\sqrt{64-64a}$ exist. Is easy to check that the root only exists when the inside part is greater than 0, so $64-64a\geq 0 \implies 64\geq 64a\implies 1 \geq a$. So a can take any value from $(-\infty,1]$.

Answer 4

First we see that $8\mid b^2$ so $b=4c$ for some integer $c$. Thus we get $$c^2=2c-a\;\;\;\;\Longrightarrow \;\;\;\;c\mid a$$ So $a=cd$ for some integer $d$ and we have now $$c=2-d$$ So all solution are $(a,b)= (2d-d^2, 8-4d)$, where $d\in \mathbb{Z}$.