# If $P$ is a polynomial with $P(3)=10$ and $P(1)=1$, then why can't all the coefficients of $P$ be integers? [closed]

If $$P$$ is a polynomial with $$P(3)=10$$ and $$P(1)=1$$, then why can't all the coefficients of $$P$$ be integers?

This question was deleted for not enough details half a year ago, therefore I'm providing them. In this question specifically, I'm asking for your help to solve this problem in 8th grader way, because I'm sure that this question might help other students to understand polynomials much better, without any higher, university-level knowledge (some provided answers there are very elegant and understandable even for a 6th grader).

To mention more, right now it would be quite hypocritical to say that 'I tried < insert any theorem > but got stuck, hence I'm asking for your help'. So I'm not saying it now, instead of that just simply asking for you to undelete this question for the reasons I mentioned above.

Thank you!

• If $P(x)=ax^2+bx+c$, then $P(3)$ means $a(3)^2+b(3)+c$, right? – ericw31415 Mar 19 '18 at 22:57
• You are completely right. – thomas21 Mar 20 '18 at 8:33
• Not to spoil the mood around here, but why is this question upvoted when the upvoting caption says "this question shows research effort" when it clearly shows nothing of the like? – Pierre Arlaud Mar 20 '18 at 15:38
• @Pierre Arlaud: ...and it is not clear at all (even an 8th grade should see this): What is P(x)? – Curd Mar 20 '18 at 16:16
• – Jyrki Lahtonen Mar 21 '18 at 14:58

## 3 Answers

If $p(x)$ is a polynomial with integer coeficients then for all integers $a,b$ you have $$a-b\mid p(a)-p(b)$$

In particular you have $$3-1\mid p(3)-p(1) = 9$$

A contradiction.

• Some commenters raised doubts whether this answer is at the appropriate level given that the question asked for a grade 8 argument. That is a valid piece of criticism, but the exchange started devolving, hence the move to chat . Remember that a question can often be answered with tools from many levels of sophistication. The voters can then show their preferences. As can the original poster. – Jyrki Lahtonen Mar 21 '18 at 9:19
• @Carl But somehow I doubt this asker is an 8th grader :-) – Jyrki Lahtonen Mar 21 '18 at 14:55
• @JyrkiLahtonen You also did away with any/all comments that had anything to do with the mathematics behind the answer. And a flag was declined that brought up this issue, where there was no stated reason for declining the flag. Comments pertaining strictly to the mathematics of the answer (such as the very first comment which sought clarification) should have been retained. Oh well. – Daniel W. Farlow Mar 21 '18 at 16:08
• @DanielW.Farlow Everything is still on the chat page Jyrki linked to. – Did Mar 21 '18 at 19:20
• @Did Yes, I know. – Daniel W. Farlow Mar 21 '18 at 19:44

If the coefficients of $P$ are all integers, then $P(odd)=odd$ (that is, the value of $P$ at an odd integer is odd) if and only if an odd number of coefficients are odd. So if $P(1)=1$, there must be an odd number of odd coefficients, but if $P(3)=10$ there must be an even number of odd coefficients. This is a contradiction, so the coefficients cannot all be integers.

• By far the most elegant using only 8th-grade math (as the original title implied). – smci Mar 20 '18 at 23:09

Suppose all coefficients of $P$ are integers. Then $Q(x) := P(x+1)-1$ is also a polynomial with integer coefficients (just expand and simplify). We have $$Q(0) = P(1)-1 = 0,$$ which means the zero-th degree coefficient of $Q$ is $0$, so $Q(x)/x$ is also a polynomial with integer coefficients. In particular this implies that $Q(2)/2$ must be an integer. But $$\frac{Q(2)}{2} = \frac{P(3)-1}{2} = \frac{10-1}{2} = \frac{9}{2}.$$ Contradiction.