The professor's son.

[BMAD]

On the blackboard the mathematics professor wrote a polynomial f(x) with integer coefficients and said, "Today is my son's birthday. When his age A is substituted for x, then f(A) = A. You will note also that f(0) = P and that P is a prime number larger than A. How old is the professor's son?

[ishank ag]

His age should be 1

If we put f(x)=P(x!)-(P-1)x

Then we will always get A =1

[Rainman]

He could be 1. f(x) could then be 3-2x, P=3=f(0). Or f(x)=5-4x, or f(x)=7-6x... He could also be 2. f(x) could then be 3-x, or 5-3x... He could also be 3, or 4, or 5...

[gavinksong]

He could be 1. f(x) could then be 3-2x, P=3=f(0). Or f(x)=5-4x, or f(x)=7-6x... He could also be 2. f(x) could then be 3-x, or 5-3x... He could also be 3, or 4, or 5...

 

If he was 2, f(x) would have to be 3-x/2 or 5-3x/2, which don't have integer coefficients.

Edited March 3, 2015 by gavinksong

[Rainman]

 

He could be 1. f(x) could then be 3-2x, P=3=f(0). Or f(x)=5-4x, or f(x)=7-6x... He could also be 2. f(x) could then be 3-x, or 5-3x... He could also be 3, or 4, or 5...

 

If he was 2, f(x) would have to be 3-x/2 or 5-3x/2, which don't have integer coefficients.

 

Seems for some reason I kept thinking f(A)=1 when it should be f(A)=A. So age 2 shouldn't work, as every term in f(2) will be even except for the constant P which is odd. In fact, for any n>1 and P>n, every term in f(n) will be divisible by n except for the constant P which is not. So f(n) is not divisible by n and hence does not equal n. The professor's son must be 1 year old.