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1974 IMO Problems/Problem 6

Revision as of 15:12, 27 October 2025 by Pf02 (talk | contribs)

Problem

Let $P$ be a non-constant polynomial with integer coefficients. If $n(P)$ is the number of distinct integers $k$ such that $(P(k))^2=1,$ prove that $n(P)-\deg(P)\le2,$ where $\deg(P)$ denotes the degree of the polynomial $P.$


Solution

Lemma: Let $P(x)$ be a polynomial with integer coefficients which is not constant. Then if $P(x)$ obtains $1$ (or $-1$) as its values for at least four times then $P(x)\neq -1$ ( or $P(x)\neq 1$) for all $x$. Proof. Assume that $P(a)=P(b)=P(c)=P(d)=1$ for $a,b,c,d$ distince. Then if there's $u$ which $P(u)=-1$ then $2=P(a)-P(u)=...$ so $P(x)-P(u)-2=(x-a)(x-b)(x-c)(x-d)Q(x)$ where $Q(x)$ is a polynomial with the integer coefficients! So $-2=(u-a)(u-b)(u-c)(u-d)Q(u)$ which is impossible cause $-2$ can not presents as product of more than three distince numbers! This proved the lemma!

Back to our problem: For convinet put $m=n(P)$ and $n=\deg P$. Firstly if $n\leq 2$ then $m-n\leq2$. Assume $n\geq3$. If equation $P(x)=1$ with more than three integer points (ie.. at least $4$) then equation $(P(x))^2=1$ implies $P(x)=1$ so $m\leq n$, ie... $m-n\leq 0\leq2$. The same case for equation $P(x)=-1$. So $m\leq 6$. If $n\geq4$ then $m-n\leq 6-n\leq 2$. Now assume that $n=3$. In this case if $m\leq 5$ then $m-n\leq 2$.

So let us show that $m<6$. In fact if $m=6$ then $P(x)-1=0$ has three integers distince roots, and the same for $P(x)+1=0$. So $P(x)-1=k_1(x-a_1)(x-a_2)(x-a_3)$ and $P(x)+1=k_2(x-b_1)(x-b_2)(x-b_3)$ where $a_i$ distince and $b_j$ distince and all with $k_1,k_2$ are integers! Then $k_2(x-b_1)(x-b_2)(x-b_3)-k_1(x-a_1)(x-a_2)(x-a_3)=2$ for all $x$. So $k_1=k_2=k$. Finally, we have $2=k(a_i-b_1)(a_i-b_2)(a_i-b_3)$ for $i=1,2,3$ and because that $\pm1$ can not presents as products of three distince numbers so $k=\pm1$, we may assume $k=1$. Because $2=(-2)\cdot 1\cdot -1$ so $\{-2,1,-1\}=\{a_i-b_1,a_i-b_2,a_i-b_3\}$ This means $3a_i-(b_1+b_2+b_3)=-2+1-1=-2$. So we must have $3a_1=3a_2=3a_3$ which follows $a_1=a_2=a_3$, which contracts!. So $m\leq 5$ and we're done.

The above solution was posted and copyrighted by pluricomplex.


Remarks (added by pf02, October 2025)

$\mathbf{1.}$ The solution above is written very badly, it could use substantial editing. Also, it is incomplete. Indeed, the statement "So $m \le 6$" is not put in a proper context, and it is not properly justified.

However, the proof is understandable, and the gap can be filled in by a diligent reader.

It is interesting to note that (if we accept it as a proof) it actually proves substantially more than the problem asked. It proves the following two statements:

$\mathbf{a:}$ If one of the equations $P(x) - 1 = 0$ and $P(x) + 1 = 0$ has $\ge 4$ integer, distinct solutions, then the other has no integer solutions.

$\mathbf{b:}$ If both equations $P(x) - 1 = 0$ and $P(x) + 1 = 0$ have integer solutions then $n(P) \le 5$.

The statement of the original problem is a trivial consequence of these statements.

$\mathbf{2.}$ This problem is very easy. It is made difficult artificially by hiding the essence of the description of the distinct, integer solutions of $P(x) - 1 = 0$ and $P(x) + 1 = 0$ behind a more general sounding statement. Had it been formulated to state the essence, it would have been much less intimidating, and easier, because the essence would have pointed towards a solution.

Here is a reformulation the problem:

$\mathbf{Problem/ 2:}$ Let $P(x)$ be a polynomial of degree $d \ge 3$. Let $m_1 =$ the number of distinct integer roots of $P(x) - 1 = 0$, and $m_2 =$ the number of distinct integer roots of $P(x) + 1 = 0$. If both $m_1 \ge 1, m_2 \ge 1$ then $m = m_1 + m_2 \le 5$.

$\mathbf{Additional\ challenges:}$

$\mathbf{a:}$ With the notation of Problem 2, we have the stronger bound $m = m_1 + m_2 \le 4$.

$\mathbf{b:}$ If $d = 2$, find $P(x)$ so that there are $4$ distinct integer roots of $P(x) - 1 = 0$ and $P(x) + 1 = 0$.

$\mathbf{c:}$ If $d \ge 3$ and both $m_1 \ge 1, m_2 \ge 1$, then one of $m_1, m_2$ equals $1$ and the other equals $2$ or $3$ (in other words, we can not have $m_1 = m_2 = 2).$

$\mathbf{3.}$ Below I will prove this Problem 2 (from which the statement of the original problem follows trivially), as well as the additional challenges.

There will be some overlap of ideas with the first solution, but for the sake of making the solution self contained, I will fill in all the details.


Solution 2

[TO BE CONTINUED]

The original thread for this problem can be found here: [1]

See Also

1974 IMO (Problems) • Resources
Preceded by
Problem 5 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Last Question
All IMO Problems and Solutions
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