One equation, two variables

From the given equation:

$\begin{aligned} 2a^2+b^2-2ab & = 1 \\ a^2 + a^2 - 2ab + b^2 & = 1 \\ a^2 + (a-b)^2 & = 1 \end{aligned}$

SInce both $a^2$ and $(a-b)^2$ are $\ge 0$ , there is a solution only when either $a^2 = 1$ and $(a-b)^2=0$ or $a^2=0$ and $(a-b)^2=1$ .

When $\begin{cases} a^2 = 1 & \implies a = \begin{cases} + 1 \implies (a-b)^2 = 0 \implies b=1 \implies (a,b) = (1,1) \\ -1 \implies (a-b)^2 = 0 \implies b=-1 \implies (a,b) = (-1,-1) \end{cases} \\ a^2 = 0 & \implies a = 0 \implies (a-b)^2 = 1 \implies b = \begin{cases} + 1 & \implies (a,b) = (0, 1) \\ -1 & \implies (a,b) =(0,-1) \end{cases} \end{cases}$

Therefore, there are $\boxed 4$ solutions.

The two posted solutions are similar so here's an alternative.

Solving $b^2-2ab+2a^2-1=0$ we get $b=a \pm \sqrt{1-a^2}$

For $b$ to be an integer (and real), we need $a=-1$ , $a=0$ or $a=1$ . When $a=\pm1$ , we get one solution for $b$ ; when $a=0$ there are two solutions, for a total of $\boxed{4}$ solution pairs $(a,b)$ .

first we rewrite the equation as:

$a^{2}+(a^{2}+b^{2}-2ab)=1$

$a^{2}+(a-b)^{2}=1$

$(a-b)^{2}=1-a^{2}$

but we know that $(a-b)^{2}\geq0$ so $1-a^{2}$ has to be greater than or equal to zero then:

$1-a^{2}\geq0$ or $a^{2}\leq1$ and we can see that only $-1,0,1$ satisfy this inequality.

so by substituting the values of $a$ we get that:

when $a = 1 \Longrightarrow b=-1$

when $a=0 \Longrightarrow b=\pm1$

when $a=-1 \Longrightarrow b=-1$

so we have Four ordered pairs : $(-1,-1) , (0,-1) , (0,1) , (1,1)$