[superprismatic]

Is there a polynomial P(x) with integer coefficients such that P(1)=3 and P(3)=2?

[Guest]

Yes, any finite sequence of integers can be defined with a polynomial with itneger values, I can't remember the formula but it involved the sum of a product.

[Guest]

Turns out...

If P(x)=∑_i=0ⁿ a_i xⁱ, then P(1)=∑_i=0ⁿ a_i =3, which is congruent to 1 modulo 2. Also, P(3)=∑_i=0ⁿ a_i*3ⁱ and working modulo 2 we have P(3)=∑_i=0ⁿ a_i*1ⁱ=∑_i=0ⁿ a_i. Thus, we have that P(3) is also congruent to ∑_i=0ⁿ a_i, 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."

[Guest]

James is right,

P(x)=-3/2(x - 3) + (x - 1)

But there are an Infinite number of others.

[superprismatic]

James is right,

P(x)=-3/2(x - 3) + (x - 1)

But there are an Infinite number of others.

Ya gotta find one with integer coefficients.

[Guest]

P(x)=3x-(x^2)+1

[CaptainEd]

Aparichit, your P(3) = 1, but the OP requires that P(3)=2.

Superprismatic, I enjoyed the "foraging", as I manually explored the landscape around these integer coefficients. I would have posted a parity-based answer, but stigge posted a clearer answer sooner. But thank you for the journey!

[Guest]

p(x)=-x/2+7/2