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So here is the Question :-

Find all $(x,y)$ pairs : $x,y$ $\in \mathbb{Z}$ such that :- $$x^4 - 4x^3 - 19x^2 + 46x = y^2 - 120.$$

What I tried :- I factored the LHS and got as :- $$x(x - 2)(x^2 - 2x - 23) = y^2 - 120$$ From here I don't know how to proceed . I can see that $(y^2 - 120)$ has $3$ factors to be broken into , and each of $x,(x - 2) , (x^2 - 2x - 23)$ divides $y^2 - 120$ , but how will I proceed from here?

Any hints or answers to this problem will be greatly appreciated !!

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    $\begingroup$ Hint. $x^4 - 4x^3 - 19x^2 + 46x+120=(x + 2) (x + 3) (x - 4) (x - 5)$ $\endgroup$ Jun 22 '20 at 5:16
  • $\begingroup$ By taking $120$ on left side, I've found the factors of LHS as $x=-2, -3$ by hit and trial method and $x=4, 5$ by further factorisation. $\endgroup$
    – SarGe
    Jun 22 '20 at 5:17
  • $\begingroup$ Ok, so from here I get $(x + 2)(x + 3)(x - 4)(x - 5)$ = $y^2$ . $\endgroup$ Jun 22 '20 at 5:22
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Here's a solution using what I think is an underappreciated problem solving technique. Note that completing the square (!) shows that $$ x^4 - 4x^3 - 19x^2 + 46x + 120 \quad \text{is close to} \quad ( x^2 - 2x - 11.5 )^2. $$ It's not hard then to show that \begin{align*} ( x^2 - 2x - 11 )^2 - (x^4 - 4x^3 - 19x^2 + 46x + 120) &= x^2-2x+1 > 0 \text{ for } x\ne1, \\ ( x^2 - 2x - 12 )^2 - (x^4 - 4x^3 - 19x^2 + 46x + 120) &= -x^2+2x+24 < 0 \text{ for } x\notin[-4,6]. \end{align*} In particular, $x^4 - 4x^3 - 19x^2 + 46x + 120$ is between the squares of two consecutive integers (and therefore cannot itself be the square of an integer) when $x\notin[-4,6]$.

It is a simple matter to check all values $-4,-3,\dots,6$ to find that $x=\{-4,-3,-2,1,4,5,6\}$ are the only values that make $x^4 - 4x^3 - 19x^2 + 46x + 120$ a perfect square. (Indeed, the symmetry around $x=1$ would reduce the amount of checking here, if we notice that $x^4 - 4x^3 - 19x^2 + 46x + 120$ is invariant under changing $x$ to $1-x$.)

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  • $\begingroup$ The “completing the square” method is done by matching the coefficients of the largest three powers of x, am I correct? $\endgroup$ Jun 22 '20 at 5:54
  • $\begingroup$ Also, how do you “notice” the invariance under $x \mapsto 1-x$? What gave you the clue? $\endgroup$ Jun 22 '20 at 5:55
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    $\begingroup$ You're correct! The clue for the invariance was that $x^2-2x$ has that invariance, and $x^2-2x$ was showing up all over the place in the identities. (Indeed, the displayed pair of equations is actually a proof that the original polynomial has that invariance, without expanding it out.) $\endgroup$ Jun 22 '20 at 6:29
  • $\begingroup$ as far as the first step, without yet being sure what is going on, we can replace $x=t+1$ to get $t^4 - 25 t^2 + 144$ which is near $(t^2 - 12.5)^2$ $\endgroup$
    – Will Jagy
    Jun 22 '20 at 19:08
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Suppose $x,y$ are integers such that $x^4-4x^3-19x^2+46x+120=y^2$.

Let $g=2x^2-4x-23$.

Identically we have $$(2x^2-4x-23)^2-4(x^4-4x^3-19x^2+46x+120)=49$$ hence $$g^2-4y^2=49$$ or equivalently $$(g+2y)(g-2y)=49$$ The average of the factors on the left is $g$.

If $49$ is expressed as the product of two integers, the average of the two factors must be an element of the set $$\{-25,-7,7,25\}$$

Thus, to find qualifying values of $x$, it remains to solve each of the equations \begin{align*} 2x^2-4x-23&=-25\\[4pt] 2x^2-4x-23&=-7\\[4pt] 2x^2-4x-23&=7\\[4pt] 2x^2-4x-23&=25\\[4pt] \end{align*} for $x$.

I'll leave the rest to you.

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