Qs. For a natural no. '$b$', let $N(b)$ denotes the no. of natural numbers '$a$' for which the equation $x^2+ax+b =0$ has integer roots. What is the smallest value of $b$ for which $N(b) =20$?
Now, by applying Sreedhar Acharya's formula, $x=[-a +- \sqrt\frac{(a^2-4b)]}{2....(i)}$ So for '$x$' being integer(as integer roots are required), the numerator part of $(i)$ is to be "even" as the denominator being $2$, any odd number would end up giving a fractional value.
So the main objective is to find the possible value of $ \ 'b' $ for which $[-a +- \sqrt{(a^2-4b)}]$ is even no. For the above quantity to be even $-a$ and $\sqrt{(a^2-4b)}$ both has to be $i$) even or both $ii$) odd.
Now as the quantity "$4b$" in $\sqrt{(a^2-4b)}$ is always even, we can only concentrate on "a" being even or odd (as if a is odd the value under the root is also odd and if even the value under root being even resulting in the whole numerator to be even in both cases) So the condition to be imposed on the root is; $$a^2-4b=c^2$$ (where $c$ is any arbitrary natural number) Now we are to solve for the minimum value of "$b$" such that "$a$" has 20 solutions $[a,b=1,2,3,4,..]$ ( since we're given that $N(b)=20$ )
But here I'm stuck in finding the solution.
Other than "trial and error method", is there any other way of solving for "$b$"?
Thank you