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Let $p,q \in \mathbb R$ and $m,n \in \mathbb N$.

Assume that $$ (x+1)^p(x-1)^q=(x+1)^m (x-1)^n \textrm{ for each } x >1. $$ Is then $p=m$ and $q=n$?

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We have $$(x+1)^{p-m}=(x-1)^{n-q}$$ But $(x+1)-(x-1)=2\implies$gcd $(x+1,x-1)|2$ – lab bhattacharjee Mar 17 '14 at 17:03
$gcd$ is for integers. We don't know if $p-m, n-q$ are natural numbers. – Alex Mar 17 '14 at 17:10
Hint: set $x=2$. – Macavity Mar 17 '14 at 18:20
up vote 6 down vote accepted

I don't understand the solutions using gcd, since these are not necessarily polynomials.

Here is the approach I would take: note that$(x+1)^{p-m}=(x-1)^{n-q}$, so we may assume that $m=n=0$. Then the question is: if $(x+1)^p (x-1)^q = 1$ for $x>1$, must we have $p=q=0$?

We can rewrite this as $(\frac{x+1}{x-1})^p (x-1)^{p+q}=1$. By letting $x$ approach $\infty$, the first term approaches $1$. If $p+q<0$, the second term approaches $0$. If $p+q>0$, the second term approaches $\infty$. It follows that $p+q=0$.

So we have $(\frac{x+1}{x-1})^p=1$. Plugging in, say, $x=2$, this can only hold if $p=0$. We conclude that $p=q=0$.

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The powers are natural numbers. If equality holds for real $x\gt 1$ it certainly holds for integers greater than $1$ - so we can choose to work with $x$ integral. – Mark Bennet Mar 17 '14 at 17:24
@MarkBennet only two of the powers are natural numbers. – Slade Mar 17 '14 at 20:07
Apologies, my brain switched off. – Mark Bennet Mar 17 '14 at 20:10
@MarkBennet Well, the problem does have "polynomials" in the title... – Slade Mar 18 '14 at 4:38

Let us make this simple.

First set $x=2\gt 1$ so equality holds. This gives us $$3^p\cdot1=3^m\cdot 1$$ from which we deduce immediately that $p=m$.

Now cancel the $(x+1)$ term from each side, noting that it cannot be zero (the powers are the same), to give $$(x-1)^q=(x-1)^n$$

It follows easily that $q=n$, for example by taking $x=3$.

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Suppose $q < n$ and $x \notin \{-1,1\}$. Divide both sides by $(x-1)^q$ to get

$(x+1)^p = (x+1)^m (x-1)^{n-q}$. Now take limits $x\to 1$ which gives a contradiction. Hence $q=n$. Similarly for $p$ and $m$.

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