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Recently, I was asked the following question by a friend : find all $a,b,c,a',b',c',k \in {\mathbb Z}$ with $k\neq 0$ such that the identity

$$ (X-a)(X-b)(X-c)+k=(X-a')(X-b')(X-c') $$

holds in ${\mathbb Z}[X]$. Does anyone have an idea ? Note that there is a certain translation invariance for the solutions : $(a,b,c,a',b',c',k)$ is a solution iff $(a+m,b+m,c+m,a'+m,b'+m,c'+m,k)$ is, for any $m\in{\mathbb Z}$.

Ewan Delanoy
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  • I think it is a difference of two cubic equation with integer roots which is equal to an integer. So, whenever $X$ is an integer, isn't this equation satisfied? Please tell me if I misunderstood the question. – Swadhin Dec 21 '14 at 07:41
  • Certainly it can't be "only a few", because if $p(X) = (X-a)(X-b)(X-c)$ is a solution (for a given $k$), then so is $p(X + m)$ for any integer $m$. So either no solutions or infinitely many. – Robert Israel Dec 21 '14 at 07:56
  • there are unknown variables? If only unknown $X$ brackets cut and get a quadratic equation. Knowing the rest will get the answer. – individ Dec 21 '14 at 07:57

1 Answers1

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These are not all solutions, but one infinite family. For example, for any integers $x,y,z$, take $a = xyz$, $b = -x(x+y)z$, $c = -y(x+y)z$, $k = x^2 y^2 (x+y)^2 z^3$. Note that $ab + ac + bc = 0$. If $p(X) = (X-a)(X-b)(X-c)$, we have $p(0) = -x^2 y^2 (x+y)^2 z^3$ and $p'(0) = 0$, and then $$p(X) + k = X^2 (X + (x^2 + xy + y^2) z )$$

EDIT: The same value of $(x^2 + x y + y^2) z$ can occur for different $x,y,z$, so this produces different solutions with $a,b,c$ distinct. For example, try $x,y,z = -3,8,1$ and $x', y', z' = -3,1,7$: we get $$ (X+24)(X-15)(X+40) = (X+21)(X+42)(X-14)-2052$$

EDIT: Let $q(a,b,c) = (X-a)(X-b)(X-c)$, so you want to solve $q(a,b,c) +k = q(a',b',c')$. Note that this is true for some $k$ iff $\dfrac{d}{dX} q(a,b,c) = \dfrac{d}{dX} q(a',b',c')$, and that is equivalent to $$ \eqalign{a + b + c &= a' + b' + c'\cr ab + ac + bc &= a'b' + a'c' + b'c'\cr}$$ Now given two integers $u,v$, we may try to solve $$ \eqalign{a + b + c &= u\cr ab + ac + bc &= v\cr}$$ Eliminating $c$ gives us $$ 3 (2a + b - u)^2 + (3 b - u)^2 = {4} u^2 - 12 v $$ Given $u$ and $v$, the equation $3 y^2 + z^2 = 4 u^2 - 12 v$ has finitely many integer solutions $(y,z)$. We want solutions where $z \equiv -u \mod 3$ (so $b = (z+u)/3$ is an integer), and where $y \equiv u - b \mod 2$ (so $a = (y+u-b)/2$ is an integer).

For example, with $u=79$ and $v = 1920$ there are $24$ integer solutions of $3 y^2 + z^2 = 1924$. Of these, $12$ satisfy $z \equiv - u \mod 3$, and all these satisfy $y \equiv u - b \mod 2$. The corresponding $(a,b,c)$ values are the six permutations of $(15,24,40)$ and the six permutations of $(12,31,36)$. Thus these correspond to $$ (X - 15)(X - 24)(X - 40) + k = (X - 12)(X - 31)(X - 36)$$ where in this case $k = 15\cdot 24 \cdot 40 - 12 \cdot 31 \cdot 36 = 1008$.

The next step would be to look at this in terms of factorization in $\mathbb Z[\sqrt{-3}]$.

Robert Israel
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