9

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?

Zev Chonoles
  • 129,973
bodacydo
  • 3,922

2 Answers2

11

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 $p(0)=10$.

Jyrki Lahtonen
  • 133,153
  • Interesting. Can you share how you managed to construct this polynomial? – bodacydo Nov 27 '11 at 20:21
  • The trick is that all the numbers $(e-a), (e-b), (e-c), (e-d)$ must be relatively small integers, if their product is to be equal to six. – Jyrki Lahtonen Nov 27 '11 at 20:22
  • Thanks for your answer. However I can't understand why it has "to be equal to six"? – bodacydo Nov 27 '11 at 20:31
  • 2
    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. – Jyrki Lahtonen Nov 27 '11 at 20:36
2

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!)

Community
  • 1
Bill Dubuque
  • 272,048
  • +1 You wouldn't mind sharing that problem? Or did it become severely spoiled with the hints you just gave :-) – Jyrki Lahtonen Nov 27 '11 at 20:33
  • @Jyrki Alas, it was far too long ago to recall the details. The idea was to compose a problem where the polynomial form was not obvious. My fuzzy memory tells me that I may have employed an Erhart polynomial. – Bill Dubuque Nov 27 '11 at 20:58
  • Now I tried another problem, where p(a)=p(b)=p(c)=p(d)=5 and p(e)=8. So using Factor Theorem, I write: a-e | p(a) - p(e) = -3, similarly, b-e | -3, c-e | -3, d-e | -3. So I have system of equations: a-e|-3, b-e|-3, c-e|-3, d-e|-3. Now there are only 4 numbers that divide -3, they are: {-1, 1, -3, 3}. So a-e = -1, b-e = 1, c-e = -3, d-e = 3. But now I am stuck. Can anyone help? – bodacydo Nov 27 '11 at 21:34
  • 1
    @bodacydo, you're missing the point that the product of the four numbers must be 3. – Gerry Myerson Nov 27 '11 at 22:23
  • @GerryMyerson I spent last 2 weeks trying to understand why the product has to be 3, but unsuccessfully. Can you give me a hint? Because in my last comment I can simply choose a to be any number, and then the system of equations {a-e = -1, b-e = 1, c-e = -3, d-e = 3} has a solution. Very stuck. Thanks. – bodacydo Dec 11 '11 at 23:25
  • 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. – Gerry Myerson Dec 12 '11 at 00:36