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I am trying to prove the following problem:

Prove that there's no such polynomial $p(x)$ with integer coefficients, such that $p(a) = p(b) = $ $p(c) = p(d) = 4$ and $p(e) = 10$, where $a, b, c, d, e$ are integers themselves and are distinct.

If $p(x)$ is the polynomial and $p(a) = p(b) = p(c) = p(d) = 4$ then it has the form $p(x) = (x-a)(x-b)(x-c)(x-d)+4.$

Now I can't figure out how I prove that there is no such integer $e$ that $p(e) = 10$.

I tried constructing various polynomials in Mathematica (as Lagrange Interpolating polynomial) and I always ended up having something like $p(x) = \frac{(x-a)(x-b)(x-c)(x-d)}{\text{const}}+4$ for a polynomial that interpolates points $(a,4), (b,4), (c,4), (d,4), (e,10)$.

I can't find a good argument that the polynomial $(x-a)(x-b)(x-c)(x-d)+4$ will always be divided by some $\text{const}$ to meet the $p(e)=10$ requirement, therefore there is no such polynomial with integer coefficients that.

Can anyone help?

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    No, I'm afraid I can't. :P2011-11-27

2 Answers 2

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The problem is that in general the factors $(e-a), (e-b), (e-c), (e-d)$ are too large. But if we are careful, we can find examples, where this happens. For example, let $a=-1$, $b=-2$, $c=1$, $d=3$, $e=0$. Then $ p(x)=(x+1)(x+2)(x-1)(x-3)+4 $ has the values $p(-1)=p(-2)=p(1)=p(3)=4$ and $p(0)=10$.

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    It cannot be three (the product of four distinct non-zero integers has absolute value at least 4). It has to be a factor of $6=10-4$, because otherwise the polynomial won't have integer coefficients.2011-11-27
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Hint $\ $ The key to proving that many problems of this type are unsolvable is to simply apply the Factor Theorem $\rm\ x-y\ |\ p(x)-p(y)\ $ in $\rm\:\mathbb Z[x,y]\ $ for $\rm\:p(x)\in \mathbb Z[x]\:.\:$ Specializing $\rm\ x,y = m,n\in\mathbb Z\:$ we deduce that $\rm\: m-n\ |\ p(m)-p(n)\:$ in $\rm\:\mathbb Z\:.\:$

For example, considering the specific example in Jyrki's answer, since $\rm\:p(a) = 4\:$ for $\rm\:a\ne 0\:$ we infer that $\rm\: a-0\ |\ p(a)-p(0) = 4-10\:,\:$ i.e. $\rm\: a\:|\:6\:$ for $\rm\:a\ne 0\:.\:$ These severe arithmetic constraints are enough to resolve the problem in many cases. A long time ago I once crafted a problem based on this combined with Pick's theorem that went unsolved for a long time till someone noticed the trick (it was John H. Conway if memory serves correct - it's not easy to pull the wool over his eyes!)

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    You've got $p(x)-5=(x-a)(x-b)(x-c)(x-d)Q$ for some integer $Q$. Now put in $x=e$ to get $3=(e-a)(e-b)(e-c)(e-d)Q$, so the product of the four numbers must be $3$ (or $-3$, or $1$, or $-1$). My apologies if, by forgetting about the possible existence of the integer multiplier $Q$, I made things more difficult than they ought to have been.2011-12-12