The website BrainDen.com offers Brain Teasers, that often tease my mind. On September 8, 2011, it posed the following problem.
Is there a polynomial P(x) whose coefficients are all integers such that P(1)=3 and P(3)=2?
Simple enough, right...now get to work. Yes or No?
Source: BrainDen.com, 9/8/2011
Hint: Suppose P(x) was linear: Ax+B = 0...will that work? Think...what is the equation of a straight line passing the points (1,3) and (3,2)?
Now focus on P(x) being quadratic: Ax2+Bx+C = 0...will that work? etc.
Solution Commentary: Some initial reasoning: The linear case is quickly eliminated, as it forces A to have the value -1/2.
Now, for the quadratic case, P(x) = Ax2+Bx+C = 0. Substitute, to find that P(1) = A+B+C = 2 and P(3) = 9A+3B+C = 2. Solving this system, we find that 8A+2B = -1, which can never happen as the left-side is even and the right-side is odd.
Now, try the cubic....is a pattern developing that can be generalized?
When the problem was posted, "Stigge" responded with more "official" reasoning: "If P(x)=∑i=0n ai xi, then P(1)=∑i=0n a_i =3, which is congruent to 1 modulo 2. Also, P(3)=∑i=0n ai*3i and working modulo 2 we have P(3)=∑i=0n ai*1i=∑i=0n ai. Thus, we have that P(3) is also congruent to ∑i=0n ai, which we knew was 1 but the condition that P(3)=2 says that P(3) is also congruent to 0 modulo 2. This is impossible, which means that there are no integers that can do the trick.
In other words, we've effectively shown that the total of these integral coefficients would have to be both odd and even at the same time... a hard feat for any integer!
In case things don't show up correctly, " ∑i=0n " is to be read "summation from i equals 0 to n."