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?
+118608 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
