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I have a problem with the following task.

Let $W(n) := an^2 + bn + c$ where $a,b,c \in \mathbb{Z}$.
Assume that for all $n \in \mathbb Z$ we have that $W(n)$ is the square of an integer.

Show that there exists some $P$ such that $W(n) = P(n)^2$.

Thanks for any tips or help.

John

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    In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post.2012-10-14
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    @JulianKuelshammer: Don't be too harsh. Obviously the "Show that" is part of a verbatim quotation introduced by the first introductory sentence. But I agree on the other points.2012-10-14
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    Was that "harsh"?2012-10-14
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    @DonAntonio: No, $4x^2+4x+1=(2x+1)^2$ is a square wnhenever $x$ is an integer, but has no *integer* root. Also (though that does not affect the problem), $W(x)=2x^2+4x+2$ has $\Delta=0$, but $W(0)=2$ is not a square.2012-10-14
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    @HagenvonEitzen: Now after the edit this is clear. +12012-10-14
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    Hagen's right: it goes only one direction.2012-10-14
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    @JulianKuelshammer: Oh, I see - I had not checked the version as it was when you commented.2012-10-14
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    for every $x$ where ??2012-10-14
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    In fact, even more is true - the result turns out to be true for polynomials of arbitrary degree and in arbitrarily many variables (and there's even an effective version that 'forces' a nonsquare value less than a specific bound for non-square polynomials). See http://www.mast.queensu.ca/~murty/poly2.pdf for the details, and a relatively straightforward proof of the univariate case.2012-10-15

3 Answers 3

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Without loss of generality, you can assume $a,b,c\geq 0$. Since

$$ 4a W(x) - (2ax+b)^2 = 4ac - b^2 = D, $$

we have that the Pell equation

$$ 4a u^2 - v^2 = D, $$

for any $N$ big enough, has at least $\left\lfloor\frac{N}{2a}\right\rfloor$ integer solutions $(u,v)$ with $|v|\leq N$. From the theory of Pell equation we know that, if $D\neq 0$, there are at most $O(\log N)$ solutions with $|v|\leq N$, so

$$ D=0, \quad a=A^2, \quad c=W(0)=C^2, \quad W(x)=(Ax+C)^2 $$

must hold.

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    Why can you exclude $D>0$? Two real roots does not imply that $W(x)$ becomes negative at an integer $x$.2012-10-14
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    Why you wrote that W(x) = u^2 ? We must prove it.2012-10-14
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    Ok, no need to exclude the case $D>0$. I wrote $W(x)=u^2$ using the fact that that $W(x)$ is a square (of an integer) for any integer value of $x$.2012-10-14
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    Ok, not a square of polynomial but square of integer?2012-10-14
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    Can you write something more about your ending of problem ? Sorry, but i cannot understand it. Why D is 0 ? Is it implicate that A is square etc. ?2012-10-14
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    If D is different from $0$, there are too many integer points on the curve $4a u^2-v^2 = D$. If $D=0$, $4au^2=v^2$, so $a$ is a square. $c$ is a square because it is $W(0)$, and if $a$ is a square, $c$ is a square and $D=0$, $W(x)$ is the square of a polynomial in $\mathbb{Z}[x]$.2012-10-14
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    Okey, thanks :)2012-10-14
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We have a (possibly nonpolynomial) function $w\colon\mathbb Z\to \mathbb N_0$ such that $W(x)=w(x)^2$ for all $x\in \mathbb Z$.

If $a<0$ then $W(x)<0$ for sufficiently big $x$. Hence $a\ge 0$. If $a=0$ and $b\ne 0$, then again $W(x)<0$ for suitable $x$. Hence $a=0$ implies $b=0$, but then $W(x)=W(0)=(w(0))^2$ as desired. Therefore we may assume for the rest of the argument that $a\ne 0$.

If $x$ is big, then $w(x)\approx x\sqrt a$. More precisely, if $\alpha,\beta$ are positive real numbers, then from $\alpha^2-\beta^2=(\alpha-\beta)(\alpha+\beta)$ we see that $|\alpha-\beta|\le\frac{|\alpha^2-\beta^2|}\alpha$. Therefore, from $(x\sqrt a+\frac{b}{2\sqrt a})^2-w(x)^2=\frac{b^2}{4a}-c$, we conclude $$\tag1w(x)=x\sqrt a +\frac b{2\sqrt a}+O(x^{-1}).$$

If $b=c=0$, then $a=w(1)^2$ and we have $W(x)=(w(1)x)^2$ as desired. Therefore we may assume that $b\ne0$ or $c\ne 0$, hence we can consider $d=\gcd(b,w(0))$ and write $b=du$, $c=d^2v^2$ with $u,v\in\mathbb Z$. Then if $p|v$, we have that $W(p)=ap^2+dup+c$ is a multiple of $p$, but not of $p^2$, contradicting squareness. Consequently, $v=1$, $d=w(0)$, $c=d^2$ and $d|b$.

As $W(\pm d)=ad^2\pm ud^2+d^2$ is divisible by $d^2$, we see that $w(\pm d)$ is divisible by $d$ and $2u=\left(\frac {w(d)}{d}\right)^2-\left(\frac{w(-d)}{d}\right)$ is the difference of two squares. But if a difference of squares is even, it is also a multiple of four. We conclude that $u$ is even. Therefore write $b=2dh$ with $h\in\mathbb Z$. Then $$\tag2W(x) = ax^2+2dhx+d^2=(hx+d)^2+(a-h^2)x^2.$$ Thus if $h^2=a$, we are done. Therefore assume that $h\ne \pm\sqrt a$. From $(2)$ we obtain $(a-h^2)x^2=(w(x)+hx+d)(w(x)-hx-d)$. Thus $(a-h^2)x^2$ has factors $w(x)+hx+d=(\sqrt a +h)x+O(1)$ and $w(x)-hx-d=(\sqrt a -h)x+O(1)$. As $x\to \infty$, both factors are unbounded, hence for a large enough prime $x$, each factor exceeds $a^2-h$ (in absolute value) and hence must be divisible by $x$ (but not $x^2$). After dividing out $x$, this implies that $a-h^2$ has integer factors $\sqrt a +h+O(x^{-1})$ and $\sqrt a -h+O(x^{-1})$. As $x$ can grow arbitrary large, $\sqrt a\pm h$ must be integer, especially $n:=\sqrt a$ is an integer. But then from $(1)$ we have $\frac{b}{2n}=w(x)-nx+O(x^{-1})$ and this implies that $m:=\frac{b}{2n}$ is an integer because $w(x)-nx$ is an integer and $O(x^{-1})$ arbitrarily small. We obtain $W(x)=(n x+m)^2+c-m^2$, hence $c-m^2 = (w(x)-nx-m)(w(x)+nx+m)$. The second factor is unbounded hence the first factor becomes arbitrarily small. But as it is an integer, this means that it becomes $0$ for large $x$. Therefore $c=m^2$ and $$W(x)=(n x+m)^2.$$

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$W(x) = ax^2 + bx + c$ is analytic and positive over the positive reals so it's square root $F(x)$ is also. Set $C$ to be the integer $F(0)$ and note that $c = C^2$. The functions $F$ and $W$ have the same set of roots but a root of $F$ is a double root of $W$ thus $b^2 = 4aC^2$ this implies some integer $A$ such that $a^2 = A$ hence $W(x) = A^2 x^2 + 2AC x + C^2 = (Ax+C)^2$.

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    Actually I think I need to extend to an analytic function on $\mathbb C$ for the existence of a root but then I'd have to mention the branch cut not getting in the way.2012-10-14
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    $W(x)=(x-1)^2$ need not be positive on the positive axis. But apparently $(x_0,\infty)$ should work just as well.2012-10-14
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    Sorry, but we must prove that W is a square of P. (and we don't know P). I think that this solution is wrong.2012-10-14
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    Yes there is a mistake with using the positive reals instead $\mathbb C$ which I hope someone else can correct.2012-10-14
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    I changed from P to F hopefully that makes it clearer.2012-10-14
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    You assume that W is square but you must prove it. You prove solution by the solution.2012-10-14
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    I use facts about the square root of W as an analytic function to show that W as a polynomial is a square.2012-10-14
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    Ok, sorry. Now i understand your solution :)2012-10-14