Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$. So I came up with $b= a+1$ $\Rightarrow$ $ab=a(a+1) = a^2 + a$
So that: 
$a^2+b^2 -1$ = $a^2 + (a+1)^2 -1$ = $2a^2 + 2a$ = $2(a^2 + a)$ $\Rightarrow$
$(a,b) = (a,a+1)$ are solutions. 
My motivation is for this follow up question:
(b) With $a$ and $b$ as above, what are the possible values of:
$$
\frac{a^2 +b^2 −1}{ab}
$$ 
Update
With Will Jagy's computations, it seems that now I must show that the ratio can be any natural number $m\ge 2$, by the proof technique of  vieta jumping. 
Update
Via Coffeemath's answer, the proof is rather elementary and does not require such technique. 
 A: As mentioned, use Vieta's root jumping technique. We know that $(1, r)$ is a solution to the problem with ratio $ \frac {1^2 + r^2 - 1}{1 \times r} = r$, but has the issue that $a = 1$ violates the condition. We can construct another solution by using the Vieta's root jumping technique.
We wish to solve $\frac {a^2 + X^2 -1 }{a \times X} = R$, which is the quadratic equation $X^2 - raX + a^2 -1 $. If this has a root $b$, then by Vieta's formula, the other root is $ra - b$ (since they sum to $r$). Note that this is independent of the condition that $a < b$.
As such, if we have any solution $(a, b)$ to the problem with ratio $R$, then we know that $( ra-b , a)$ is also going to be a solution to the problem with ratio $R$. It remains to check that $0 < ra-b < a$. 
Hence, given any solution $(a,b)$, we can, through a finite series of steps, reduce it to $(1, r)$ for some $r$, while keeping the ratio constant. This ratio is then clearly $r$.
A: If you want the ratio $(a^2+b^2-1)/(ab)$ to be equal to $a$, choose $b=a^2-1$.
Then the numerator is $$a^2+b^2-1=a^2+(a^2-1)^2-1=a^4-a^2=a^2(a^2-1),$$
which divided by the denominator $ab=a(a^2-1)$ gives the result $a$.
This answers in the affirmative the question of the "update" of the OP, since here $a$ can be any $m \ge 2$.
I noticed after posting this that Will Jagy mentions the choice $a=r,b=r^2-1$ in a comment. I'll delete this if asked to...
A: $(3,8)$ is a possible solution.
This gives us 24 divides 72, and a value of 3 for (b).
Have you considered that if $ab$ divides $a^2+b^2-1$, then we $ab$ divides $a^2 + b^2 -1 + 2ab$?
This gives us $ab$ divides $(a+b+1)(a+b-1)$.
Subsequently, the question might become easier to work with.
