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Find all values of positive integer $x$ at which the following expression is perfect square

$$P(x)=x^4-4x^3+22x^2-36x+18$$

I tried to assume $P= (x^2+ax+b)^2$ ; and comparing the cofactors , get that $a= -2 ; b= 9$ , but when expand the $x^2-2x+9$, I didn't get the same as $P$

What is wrong in my work?

user373141
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    In the statement of the problem, you neglected to specify the restrictions on $x$. – quasi Aug 24 '17 at 20:23
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    Why do you assume that the given polynomial can be written as a perfect square? There are plenty of x values that do not produce a function value that are squares. For example, x=0 gives 18. It is impossible for $x$ to be any random value – imranfat Aug 24 '17 at 20:23
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    Such problems can be difficult to solve, and your approach is not quite up to the task. You are hoping to show the polynomial $P(x)$ is the square of some (quadratic) polynomial, but if that were so, then every integer $x$ would make $P(x)$ a perfect square. As @imranfat points out, this is just not so. The problem is to find integer (or positive integer, if your last Comment is correct) solutions to $y^2 = x^4 -4x^3-22x^2-36x+18$. It is possible that factoring $P(x)$ will help with this Diophantine problem. – hardmath Aug 24 '17 at 20:30
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    Btw this polynomial is irreducible, let alone being a perfect square(for every $x$). – Vidyanshu Mishra Aug 24 '17 at 20:32
  • It is possible that reducing mod $p$ may help a bit; for instance, reducing mod $3$ shows $x$ must be congruent to $0$ or $2$ mod $3$. Not sure how helpful this will be though – TomGrubb Aug 24 '17 at 20:32
  • Reducing mod 4 shows that $x$ must be odd. – Gribouillis Aug 24 '17 at 20:34
  • And mod 5 gives $x\equiv3\pmod5$. – Jyrki Lahtonen Aug 24 '17 at 20:36
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    For sufficiently large $x$, we have $ (x^2 - 2 x - 14)^2\le P (x)\le (x^2 - 2 x - 2)^2$. –  Aug 24 '17 at 20:40

2 Answers2

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As Jyrki Lahtonen mentioned in comments, my bounds can be improved to

$$(x^2-2x-14)^2 < P(x) < (x^2-2x-13)^2,$$

or in expanded form:

$$196 + 56 x - 24 x^2 - 4 x^3 + x^4 < 18 - 36 x - 22 x^2 - 4 x^3 + x^4 < 169 + 52 x - 22 x^2 - 4 x^3 + x^4$$

for sufficiently large values of $x$ ($x \ge 48$ in first inequality, $x \ge -1$ in the second one). But these bounds are actually squares of adjacent integers, so after inspection of finitely many cases ($x = 1, 2, \ldots, 47$) we can surelz say that $P(x)$ is never a square.

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If $x$ is an integer, then $$x^4-4x^3-22x^2-36x+18$$ $\qquad$is congruent to $2$ mod $4$ if $x$ is even,

$\qquad$and is congruent to $5$ mod $8$ if $x$ is odd,

hence can't be a perfect square.

(Even squares are congruent to $0$ mod $4$, and odd squares are congruent to $1$ mod $8$.)

Some details . . .

If $x$ is even, all the terms are multiples of $4$ except $18$, so \begin{align*} &x^4-4x^3-22x^2-36x+18\\[4pt] \equiv\;&18\;(\text{mod}\;4)\\[4pt] \equiv\;&2\;(\text{mod}\;4)\\[4pt] \end{align*} If $x$ is odd, then $x^2 \equiv 1\;(\text{mod}\;8)$, hence \begin{align*} &x^4-4x^3-22x^2-36x+18\\[4pt] \equiv\;&x^4-4x^3-6x^2-4x+2\;(\text{mod}\;8)\\[4pt] \equiv\;&(x^2)^2-4(x^2)x-6x^2-4x+2\;(\text{mod}\;8)\\[4pt] \equiv\;&(1)^2-4(1)x-6(1)-4x+2\;(\text{mod}\;8)\\[4pt] \equiv\;&-3-8x\;(\text{mod}\;8)\\[4pt] \equiv\;&5\;(\text{mod}\;8)\\[4pt] \end{align*}

quasi
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