Problem:
Let P(x) be a polynomial with integer coefficients that satisfies P(17)=10 and P(24)=17. Given that the equation P(n)=n+3 has two distinct integer solutions n1 and n2, find the product n1⋅n2.
Solution:
Let S(x)=P(x)−x−3. Because S(17)=−10 and S(24)=−10,
S(x)=−10+(x−17)(x−24)Q(x)
for some polynomial Q(x) with integer coefficients. If n is an integer such that P(n)=n+3, then S(n)=0, and (n−17)(n−24)Q(n)=10. Thus the integers n−17 and n−24 are divisors of 10 that differ by 7. The only such pairs are (2,−5) and (5,−2). This yields {n1,n2}={19,22}, hence n1⋅n2=418. An example of a polynomial that satisfies the conditions of the problem is P(x)= x−7−(x−17)(x−24).
The problems on this page are the property of the MAA's American Mathematics Competitions