Irish Mathematical Olympiad

a.
$$(x^2\sqrt{2})^2 + 8 = 2(x^4+4) = 2((x^2+2)^2 - 4x^2) = 2(x^2-2x+2)(x^2+2x+2).$$

b.
$g(x) = x^4 + x^3 + x^2 + x + 1 = (x - \zeta)(x - \zeta^2)(x - \zeta^3)(x - \zeta^4)$, where $\zeta$ is a primitive fifth root of unity. Using $\zeta = \zeta^6$ and $\zeta^3 = \zeta^8$ we have
$$g(x) = (x - \zeta^2)(x - \zeta^4)(x - \zeta^6)(x - \zeta^8).$$
As $5x^2 - \zeta^{2k} = (x\sqrt{5} + \zeta^k)(x\sqrt{5} - \zeta^k)$, we can write $g(5x^2) = g_1(x)g_2(x)$ where
$$g_1(x) = (x\sqrt{5} + \zeta)(x\sqrt{5} + \zeta^4)(x\sqrt{5} - \zeta^2)(x\sqrt{5} - \zeta^3) \text{ and}$$
$$g_2(x) = (x\sqrt{5} + \zeta^2)(x\sqrt{5} + \zeta^3)(x\sqrt{5} - \zeta)(x\sqrt{5} - \zeta^4).$$

When multiplying these out, we are interested in $z = \zeta + \zeta^4 = \zeta + \zeta^{-1}$. Using $g(\zeta) = 0$ we easily verify that $z^2 + z - 1 = 0$. Therefore $z = (-1 \pm \sqrt{5})/2$ and $2z + 1 = \pm\sqrt{5}$. Moreover, $\zeta^2 + \zeta^3 = \zeta^2 + \zeta^{-2} = z^2 - 2 = -(z + 1)$. Hence
$$\begin{aligned} g_1(x) &= (5x^2 + xz\sqrt{5} + 1)(5x^2 + x(z+1)\sqrt{5} + 1) \\ &= 25x^4 + 5\sqrt{5}(2z+1)x^3 + 5(z^2+z+2)x^2 + \sqrt{5}(2z+1)x + 1 \\ &= 25x^4 \pm 25x^3 + 15x^2 \pm 5x + 1. \end{aligned}$$
This gives the factorisation
$$g(5x^2) = (25x^4 + 25x^3 + 15x^2 + 5x + 1)(25x^4 - 25x^3 + 15x^2 - 5x + 1).$$

c.
$h(x) = (x+1)^{13} - x^{13} - 1$ has obvious roots $x = 0, x = -1$. A key observation is that if $\omega$ is a primitive cube root of unity, then $\omega + 1 = -\omega^2$, so $(\omega+1)^{13} - \omega^{13} - 1 = -\omega^2 - \omega - 1 = 0$. Since $(x-\omega)(x-\omega^2) = x^2+x+1$, this gives the factor $x(x+1)(x^2+x+1)$ of $h(x)$. By long division, we obtain $h(x) = 13x(x+1)(x^2+x+1)k(x)$, where
$$k(x) = x^8 + 4x^7 + 12x^6 + 22x^5 + 27x^4 + 22x^3 + 12x^2 + 4x + 1.$$
We find $k(\omega) = 0$ and thus $k(x)$ is divisible by $x^2 + x + 1$, and the result is proved.
Alternatively, we might use that the derivative $h'(x) = 13(x+1)^{12} - 13x^{12}$ has $\omega$ as a root, which is easily seen. This implies that $h(x)$ has a root of order at least two at $\omega$. Without long division we see now that $h(x)$ has the factor of degree six $x(x+1)(x^2+x+1)^2$ and the result follows again. The factorisation of $h(x)$ explicitly is
$$13(x^6 + 3x^5 + 5x^4 + 5x^3 + 3x^2 + x)(x^6 + 3x^5 + 8x^4 + 11x^3 + 8x^2 + 3x + 1).$$