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MinhDucDangCHL2000

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MinhDucDangCHL2000

#1 h Apr 16, 2025, 2:44 PM • 1 Y

Y by Mhuy

Let $P(x)$ be a polynomial with integer coefficients. Suppose there exist infinitely many integer pairs $(a,b)$ such that $P(a) + P(b) = 0$ . Prove that the graph of $P(x)$ is symmetric about a point (i.e., it has a center of symmetry).

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#3 h Apr 16, 2025, 4:12 PM • 3 Y

Y by MinhDucDangCHL2000, Mhuy, ZeltaQN2008

Nice problem!

Let's plot the graph of a polynomial function $y=P(x)$ on the Cartesian coordinate plane. If it has a center of symmetry, we can translate the graph so the center moves to the origin $O(0,0)$ , and the resulting function becomes an odd function. This means the graph of $P(x)$ is symmetric about a point $(\alpha, \beta)$ if and only if the translated function $Q(x) = P(x+\alpha) - \beta$ is an odd function, i.e., $P(x+\alpha) + P(-x+\alpha) = 2\beta$ holds for some fixed $\alpha, \beta$ and for all $x$ .

Now, it is easy to see that a polynomial satisfying the problem's given condition must be of odd degree. We can also assume it is monic:
$P(x)= x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0.$ Thus, we can fix some integers $a,b\in \mathbb Z$ with $a>0, b<0$ such that $P(a)=P(b)$ , and the function $P(x)$ increases on $(a,+\infty)$ and decreases on $(-\infty,b)$ . Without loss of generality (WLOG), assume $a\geq-b$ . Let $t=a+b+1$ . Since $a\ge -b$ and $a>0$ , we have $a+b \ge 0$ , so $t \ge 1 > 0$ .

Claim: $P(a+m+t)>-P(b-m)$ for all sufficiently large $m$ .

Claim proof: The inequality $P(a+m+t)>-P(b-m)$ is equivalent to $[n(a+t)+1]m^{n-1}+\text{lower order terms in } m > [-nb+1]m^{n-1} +\text{lower order terms in} \;m.$ This is always true for sufficiently large $m$ . $\square$

Now, assume there are infinitely many pairs of sufficiently large positive integers $(h,k)$ such that $P(a+h)=-P(b-k)$ . From the claim, we must have $h < k+t$ , or $h-k < t$ . Similarly, one can argue that $h-k > -t$ . This means that for these infinitely many pairs $(h,k)$ , the integer difference $(h-k)$ must lie in the finite interval $[-t, t]$ . By the Pigeonhole Principle (PHP), since there are infinitely many such pairs $(h,k)$ but only a finite number of integer values in $[-t, t]$ , there must be infinitely many pairs $(h,k)$ satisfying the condition $P(a+h)=-P(b-k)$ for which $h-k=q$ for some fixed integer $q \in [-t, t]$ .

Therefore, there are infinitely many integers $h$ such that
$P(a+h)=-P(b-h+q).$ By the identity theorem for polynomials, this implies the polynomial identity:
$P(a+x) \equiv -P(-x+b+q).$ Let $\alpha = \frac{a+b+q}{2}$ . Replacing $x$ with $x-\frac{a-b-q}{2}$ in the identity gives:
$P\left(x+\frac{a+b+q}{2}\right) = -P\left(-x+\frac{a+b+q}{2}\right).$ This equation signifies that the function $Q(x) = P\left(x+\frac{a+b+q}{2}\right)$ is an odd function ( $Q(x)=-Q(-x)$ ), which means the graph of the original polynomial $P(x)$ has a center of symmetry.

This post has been edited 1 time. Last edited by Tung-CHL, Apr 16, 2025, 4:18 PM
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