Help needed

The polynomial p(x) = x^2 + ax + b has positive integer coefficients a and b. If p(60) is a perfect
square and the equation p(x) = 0 has two distinct integer solutions, what is the least possible value of b?

p(x) = 0 has two distinct integer solutions
$a^2-4b  \text{  is a perfect square}\\ a^2(1-\frac{4b}{a^2}) \text{  is a perfect square}\\ (1-\frac{4b}{a^2}) \text{  is a perfect square}\\ \text{but a and b are both positive so 4b must equal }a^2\\ 4b=a^2\\ \text{maybe b=1 and a=2}$

p(60)=3600+60a+b

sub in b=1 and a=2

3600+120+1 = (60+1)^2=61^2    so that works perfectly.

The least possible value of b is 1