3

I tried to find the value of integers $a$, $b$, $c$, $d$, $w$, $y$, and $z$ so that the polynomial
$(x+a)(x+b)(x+c)+d$ and $(x+w)(x+y)(x+z)$ are equal.

By using brute-force approach with Microsoft Excel, I find some solutions like:

$(x-13)(x-12)(x-8) - 12 = (x-9)(x-10)(x-14)$

$(x-13)(x-10)(x-10) + 4 = (x-12)(x-9)(x-12)$

$(x+6)(x+7)(x+11) - 12 = (x+9)(x+10)(x+5)$

$(x-7)(x+3)(x-13) - 288 = (x-1)(x-15)(x-1)$

and so on.

To my surprise, all the values of $d$ that I obtained are multiples of $4$.

Is it true that $d$ is always a multiple of $4$? If so, why?

I tried to use Wolfram Alpha to find out why this happens but it gave the solutions $w$, $y$, and $z$ in terms of $a$, $b$, $c$, and $d$ and the solutions are very long and complicated.

Note: I have also tried to find the integers $a$, $b$, $c$, $d$, $m$, $v$, $w$, $y$, and $z$ that solve the equation $(x+a)(x+b)(x+c)(x+d)+m =(x+v)(x+w)(x+y)(x+z)$ and found that $m$ is also a multiple of $4$. Maybe there is some generalization to what I found.

Negiholic
  • 33

1 Answers1

3

Indeed, every possible integral $d$ must be a multiple of $4$. To see this, first note that, for the identity to hold, we must have: $$a+b+c=w+y+z$$ and $$ab+bc+ac=wy+yz+zx$$ This means that, $$a+b+c \equiv w+ y+z \mod 4 {\tag 1}$$ and $$ab+bc +ac \equiv wz+zy+wy \mod 4 {\tag 2}$$ Thus, modulo $4$, only possible unordered triplets for $\{(a,b,c), (w,y,z)\}$ satisfying both conditions $(1)$ and $(2)$ are:

If $a+b+c \equiv 0 \mod 4$- $$\{(0,0,0), (0,2,2)\}$$

If $a+b+c \equiv 1 \mod 4$-

None

If $a+b+c \equiv 2 \mod 4$- $$\{(0,3,3), (1,1,0)\}$$ $$\{(0,2,0),(2,2,2)\}$$

If $a+b+c \equiv 3 \mod 4$- $$\{(1,1,1),(3,1,3)\}$$ $$\{(0,3,0),(3,2,2)\}$$

These pairs are those apart from the ones where $(a,b,c)=(w,y,z)$ which are of course satisfying both conditions. For each of the listed above pairs, check that $d=wyz-abc \equiv 0 \mod 4$. Hence $4|d$ is always true.

Ritam_Dasgupta
  • 5,992
  • 2
  • 8
  • 23