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Difference between revisions of "1974 IMO Problems/Problem 6"
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<math>\mathbf{2.}</math>  This problem is very easy.  It is made difficult
 
<math>\mathbf{2.}</math>  This problem is very easy.  It is made difficult
−
artificially by hiding the essence of describing the distinct,
+
artificially by hiding the essence of the description of the
−
integer solutions of <math>P(x) - 1 = 0</math> and <math>P(x) + 1 = 0</math> behind
+
distinct, integer solutions of <math>P(x) - 1 = 0</math> and <math>P(x) + 1 = 0</math>
−
a more general sounding conclusion.  Had it been formulated to
+
behind a more general sounding statement.  Had it been formulated
−
state the essence it would have been much less intimidating,
+
to state the essence, it would have been much less intimidating,
−
because the essence would have pointed towards a solution.
+
and easier, because the essence would have pointed towards a
 +
solution.
  
 
Here is a reformulation the problem:
 
Here is a reformulation the problem:
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Let <math>m_1 =</math> the number of distinct integer roots of <math>P(x) - 1 = 0</math>,
 
Let <math>m_1 =</math> the number of distinct integer roots of <math>P(x) - 1 = 0</math>,
 
and <math>m_2 =</math> the number of distinct integer roots of <math>P(x) + 1 = 0</math>.
 
and <math>m_2 =</math> the number of distinct integer roots of <math>P(x) + 1 = 0</math>.
−
If both <math>m_1 \ge 1, m_2 \ge 1</math> then <math>m_1 + m_2 \le 4</math>.
+
If both <math>m_1 \ge 1, m_2 \ge 1</math> then <math>m = m_1 + m_2 \le 4</math>.
  
 
<math>\mathbf{Additional\ challenges:}</math>  If <math>d = 2</math>, find <math>P(x)</math> so that
 
<math>\mathbf{Additional\ challenges:}</math>  If <math>d = 2</math>, find <math>P(x)</math> so that
−
there are <math>4</math> distinct integer roots of
+
there are <math>4</math> distinct integer roots of <math>P(x) - 1 = 0</math> and
−
<math>P(x) - 1 = 0</math> and <math>P(x) + 1 = 0</math>.
+
<math>P(x) + 1 = 0</math>.
  
 
If <math>d \ge 3</math> and both <math>m_1 \ge 1, m_2 \ge 1</math>, then one of <math>m_1, m_2</math>
 
If <math>d \ge 3</math> and both <math>m_1 \ge 1, m_2 \ge 1</math>, then one of <math>m_1, m_2</math>

Revision as of 01:36, 27 October 2025

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 correct, but it is written very badly. However, it is understandable, and I will refrain from rewriting it.

It is interesting to note that it actually proves substantially more than the problem asked. It proves the following two statements:

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.

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

The 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:

$\mathbf{Problem:}$ 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 $m = m_1 + m_2 \le 4$.

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

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 (from which the original problem follows easily), 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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