Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "1974 IMO Problems/Problem 6"
Line 1: Line 1:
 
==Problem==
 
==Problem==
 +
 
Let <math>P</math> be a non-constant polynomial with integer coefficients. If <math>n(P)</math> is the number of distinct integers <math>k</math> such that <math>(P(k))^2=1,</math> prove that <math>n(P)-\deg(P)\le2,</math> where <math>\deg(P)</math> denotes the degree of the polynomial <math>P.</math>
 
Let <math>P</math> be a non-constant polynomial with integer coefficients. If <math>n(P)</math> is the number of distinct integers <math>k</math> such that <math>(P(k))^2=1,</math> prove that <math>n(P)-\deg(P)\le2,</math> where <math>\deg(P)</math> denotes the degree of the polynomial <math>P.</math>
 +
  
 
==Solution==
 
==Solution==
 +
 
Lemma: Let <math>P(x)</math> be a polynomial with integer coefficients which is not constant. Then if <math>P(x)</math> obtains <math>1</math> (or <math>-1</math>) as its values for at least four times then <math>P(x)\neq -1</math> ( or <math>P(x)\neq 1</math>) for all <math>x</math>.
 
Lemma: Let <math>P(x)</math> be a polynomial with integer coefficients which is not constant. Then if <math>P(x)</math> obtains <math>1</math> (or <math>-1</math>) as its values for at least four times then <math>P(x)\neq -1</math> ( or <math>P(x)\neq 1</math>) for all <math>x</math>.
 
Proof. Assume that <math>P(a)=P(b)=P(c)=P(d)=1</math> for <math>a,b,c,d</math> distince. Then if there's <math>u</math> which <math>P(u)=-1</math> then <math>2=P(a)-P(u)=...</math> so <math>P(x)-P(u)-2=(x-a)(x-b)(x-c)(x-d)Q(x)</math> where <math>Q(x)</math> is a polynomial with the integer coefficients! So
 
Proof. Assume that <math>P(a)=P(b)=P(c)=P(d)=1</math> for <math>a,b,c,d</math> distince. Then if there's <math>u</math> which <math>P(u)=-1</math> then <math>2=P(a)-P(u)=...</math> so <math>P(x)-P(u)-2=(x-a)(x-b)(x-c)(x-d)Q(x)</math> where <math>Q(x)</math> is a polynomial with the integer coefficients! So
Line 17: Line 20:
 
This means <math>3a_i-(b_1+b_2+b_3)=-2+1-1=-2</math>. So we must have <math>3a_1=3a_2=3a_3</math> which follows <math>a_1=a_2=a_3</math>, which contracts!. So <math>m\leq 5</math> and we're done.
 
This means <math>3a_i-(b_1+b_2+b_3)=-2+1-1=-2</math>. So we must have <math>3a_1=3a_2=3a_3</math> which follows <math>a_1=a_2=a_3</math>, which contracts!. So <math>m\leq 5</math> and we're done.
  
 +
The above solution was posted and copyrighted by pluricomplex.
 +
 +
 +
==Remarks (added by pf02, October 2025)==
 +
 +
<math>\mathbf{1.}</math>  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 <math>P(x) - 1 = 0</math> and <math>P(x) + 1 = 0</math>
 +
has <math>\ge 4</math> integer, distinct solutions, then the other has
 +
no integer solutions.
 +
 +
b. If both equations <math>P(x) - 1 = 0</math> and <math>P(x) + 1 = 0</math> have
 +
integer solutions then <math>n(P) \le 5</math>.
 +
 +
The problem is a trivial consequence of these statements.
 +
 +
<math>\mathbf{2.}</math>  This problem is very easy.  It is made difficult
 +
artificially by hiding the essence of describing the distinct,
 +
integer solutions of <math>P(x) - 1 = 0</math> and <math>P(x) + 1 = 0</math> behind
 +
a more general sounding conclusion.  Had it been formulated to
 +
state the essence it would have been much less intimidating,
 +
because the essence would have pointed towards a solution.
 +
 +
Here is a reformulation the problem:
 +
 +
<math>\mathbf{Problem:}</math>  Let <math>P(x)</math> be a polynomial of degree <math>d \ge 3</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>.
 +
If both <math>m_1 \ge 1, m_2 \ge 1</math> then <math>m_1 + m_2 \le 4</math>.
 +
 +
<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
 +
<math>P(x) - 1 = 0</math> and <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>
 +
equals <math>1</math> and the other equals <math>2</math> or <math>3</math> (in other words, we can
 +
not have m_1 = m_2 = 2$).
 +
 +
 +
 +
 +
 +
 +
The original thread for this problem can be found here: [https://aops.com/community/p364890]
  
−
The above solution was posted and copyrighted by pluricomplex. The original thread for this problem can be found here: [https://aops.com/community/p364890]
 
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=1974|num-b=5|after=Last Question}}
 
{{IMO box|year=1974|num-b=5|after=Last Question}}
−
[[Category:Olympiad Geometry Problems]]
+
[[Category:Olympiad Algebra Problems]]

Revision as of 19:35, 26 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 describing the distinct, integer solutions of $P(x) - 1 = 0$ and $P(x) + 1 = 0$ behind a more general sounding conclusion. Had it been formulated to state the essence it would have been much less intimidating, 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 $m_1 + m_2 \le 4$.

$\mathbf{Additional challenges:}$ 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$).




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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1974_IMO_Problems/Problem_6&oldid=264212"