An algebra problem by mridul jain

The quartic polynomial can either be factored into a cubic and linear or 2 quadratics.

Case 1: A cubic factor and a linear factor

Let $x^4-nx+63 = (x^3+ax^2+bx+c)(x-a)$ for some integers a,b and c. We let the linear factor be $x-a$ to let the coefficient of the $x^3$ term be 0.

Its expansion is $x^4+(b-a^2)x^2+(c-ab)x-ac$ .

Comparing coefficients, we get $b = a^2$ and $ac = -63 = -3^2*7$

Note that $n = c-ab = c-a^3 > 0$ , so $c$ is positive and $a$ is negative.

$(a,c) = (-1,63),(-3,21),(-7,9),(-9,7),(-21,3),(-63,1)$

The smallest solution of n in this case is $21-(-3)^3=48$ .

Case 2: Two quadratic factors

Let $x^4-nx+63 = (x^2+ax+b)(x^2-ax+c)$ for some integers a,b and c. Again, the cubic term cancels out.

Its expansion is $x^4+(b+c-a^2)x^2+a(c-b)x+bc$ .

Comparing coefficients, we get $b+c=a^2$ and $bc=63=3^2*7$ .

In these two equations, b and c are symmetric, so we only have to look at $(b,c) = (1,63),(3,21),(7,9)$

$a = \pm \sqrt{b+c} = \pm 4$ or $\pm 8$

Hence case 2 yields the smallest solution of $n = a(c-b) = 4(9-7) = \boxed {8}$

Having done this with a symbolic algebra system (namely sympy) I can now see how others would do it in a clever way.