2017 Algebraic problem of the Year!

$\begin{aligned} y & = {\color{#3D99F6}(15-x)}{\color{#D61F06}(17-x)} {\color{#3D99F6}(15+x)}{\color{#D61F06}(17+x)} \\ & = {\color{#3D99F6}\left(225-x^2\right)}{\color{#D61F06}\left(289-x^2\right)} \\ & = x^4 - 514x^2 + 65025 \\ & = x^4 - 2(257)x^2 + 257^2 - 257^2 + 65025 \\ & = \left(x^2 - 257\right)^2 - 66049 + 65025 \\ & = \left(x^2 - 257\right)^2 - 1024 \end{aligned}$

We note that $\left(x^2 - 257\right)^2 \ge 0$ , and $\min (y) = \boxed{-1024}$ when $\left(x^2 - 257\right)^2= 0$ .

$\begin{aligned}y&=(15-x)(17-x)(15+x)(17+x)\\&=x^4-514x^2+65025\\\frac{dy}{dx}&=4x^3-1028x\\\frac{dy}{dx}&=4x^3-1028x=0\end{aligned}$

solving the equation we get, $x=0$ or, $x=\sqrt{257}$ or, $x=-\sqrt{257}$

If, $x=0$ , $y=65025$

If, $x=\sqrt{257}$ , $y=-1024$

If, $x=-\sqrt{257}$ , $y=-1024$

Hence, the minimum value of y is $\boxed{-1024}$