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I found that if $f(x)=ax^2+bx+c(a,b,c\in\mathbb{Z},a\neq0)$ always form a pefect square number for all $x\in\mathbb{Z}$,Then there must exists $s,t\in\mathbb{Z}$ so that $f(x)=(sx+t)^2$,but I cannot prove it.

Since $f(0)=c$ is a pefect square,there exists $t^2=c$. Then consider any prime $p|t$,$f(p)=ap^2+bp+t^2=n^2$,so $p|n$,$p^2|n^2$.Because $p^2|t^2$,there must be $p|b$ for all $p|t$,so $t|b$,It should be $t|b$ since I hope b=2st,but I get stuck here. What's more ,by consider $b=\frac{f(1)-f(-1)}{2}$ and perfect square module 4,I can prove b is a even number.

Bill Dubuque
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Yan GaoZhen
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2 Answers2

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This is a special case of Hilbert's irreducibility theorem, as noted explicitly in Wikipedia's article on that theorem. Specializing the proof yields the following argument, for which it is enough to assume that $f(x)$ takes square values for a few sufficiently large consecutive integer values of $x$.

For integers $n$, let $g_n = \sqrt{f(n)}$. By hypothesis each $g_n$ is an integer. On the other hand, there is some constant $C$ such that the second finite difference $d_n := g_{n+2} - 2 g_{n+1} + g_n$ satisfies $|d_n| < C/n$ for large $n$. (This can be seen in several ways, e.g. by applying Rolle's theorem twice, or by expanding $d_n$ in a Laurent series about $n = \infty$.) Since $d_n$ is an integer, it follows that $d_n = 0$ for large $n$, say $n>N_0$. But this means that $g_n$ is eventually linear: there exist integers $A,B$ such that $g_n = An+B$ for all $n > N_0$. Then $f(n) = (An+B)^2$ for all $n > N_0$, which makes $f(x) = (Ax+B)^2$ identically, QED.

Noam D. Elkies
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  • Thank you professor, and the solution is quite advanced for a K-12 math teacher. Wish could be able to digest the proof instantly. – DeepSea Aug 21 '16 at 02:35
  • Basically all I'm saying is that for large $n$ we have $g_n = a^{1/2} n + a^{-1/2} b/2 + {}$ lower order terms, so if we write $e(n) = f(n+1) - f(n)$ then $e(n) = a^{1/2} + {}$ lower order, and repeating the procedure once more we find that $e(n+1) - e(n) \to 0$ as $n \to \infty$. But $e(n+1) - e(n) = g_n$ which is an integer, so eventually all $g_n$ are zero, so all $e(n)$ are equal $-$ say $e(n) = A$ $-$ and then by induction $g_n = An+B$. – Noam D. Elkies Aug 21 '16 at 02:51
  • Thank you professor. I remembered you in the Putnam days at Harvard back in the nineties or earlier than that, and your unique approach to math is amazing . – DeepSea Aug 21 '16 at 03:00
  • Thanks. My "Putnam days" were 1982-85, and alas I did not get to spend them at Harvard, but I've been here since then. – Noam D. Elkies Aug 21 '16 at 05:12
  • Thanks for your answer and introduction of Hilbert's irreducibility theorem.But I don't understand how to "applying Rolle's theorem twice" to get$d_n<C/n$(another approach by Laurent expanding is easier to me),Could you give a more detailed explain for this? – Yan GaoZhen Aug 21 '16 at 14:15
  • More precisely I should have written "mean value theorem" instead of "Rolle's theorem". The MVT says $g_{n+1} - g_n$ is the value at some $x_n \in (n,n+1)$ of the derivative of $\sqrt{f(n)}$. Replacing $n$ by $n+1$ and subtracting, we find that $d_n$ is $x_{n+1} - x_n$ time the value somewhere in $(n,n+2)$ of the second derivative. But that second derivative is $O(1/n)$ for large $n$, and $x_{n+1} - x_n < 2$. – Noam D. Elkies Aug 21 '16 at 15:26
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    Thanks for the further explaination ,now I totally understand the proof. – Yan GaoZhen Aug 22 '16 at 09:06
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Solutions of the equation:

$$ax^2-by^2+cx-dy+q=0$$

you can record if the root of the whole: $k=\sqrt{(c-d)^2-4q(a-b)}$

Then using the solutions of the equation Pell: $p^2-abs^2=\pm1$

Then the formula of the solution, you can write:

$$x=\frac{\pm1}{2(a-b)}(((d-c)\pm{k})p^2+2(bk\mp(bc-ad))ps+b(a(d+c)-2bc\pm{ak})s^2)$$

$$y=\frac{\pm1}{2(a-b)}(((d-c)\pm{k})p^2+2(ak\mp(bc-ad))ps-a(b(d+c)-2ad\mp{bk})s^2)$$

If the root is a need to find out if this is equivalent to the quadratic form in which the root of the whole. This is usually accomplished this replacement: $x$ in such number $x+ty$

Forgot to say. The characters inside the brackets do not depend on the sign of the Pell equation.

It depends only before $\pm{1}$

individ
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