Find the largest positive integer n < 30 such that (1)/(2)(n^8+3...

Let $f(n)=\frac{1}{2}\left(n^8+3 n^4-4\right)$=\frac{1}{2}(n-1)(n+1)\left(n^2+1\right)\left(\left(n^2+1\right)+1\right)\left((n-1)^2+1\right)$For$n=2 k+1 \Rightarrow \quad f(2 k+1)=\frac{1}{2}(2 k) 2(k+1)\left(4 k^2+4 k+2\right) \ldots$Clearly$4 \mid f(2 k+1) \Rightarrow \quad$Only even cases$\Rightarrow \quad n=28, f(28)=\frac{1}{2} \times 27 \times 29 \times\left(28^2+1\right)\left(27^2+1\right)\left(29^2+1\right)$Clearly$3^2 \mid f(28) n=26, f(26)=\frac{1}{2} \times 25 \times 27 \times\left(25^2+1\right)\left(26^2+1\right)\left(27^2+1\right)$Clearly$5^2 \mid f(26) n=24, \frac{1}{2} \cdot 23 \times 25 \cdot\left(24^2+1\right)\left(25^2+1\right)\left(26^2+1\right)$Again$5^2 \mid f(26) $$\begin{aligned} & n=22, f(22)=\frac{1}{2} \times 21 \times 23 \times\left(22^2+1\right)\left(21^2+1\right)\left(23^2+1\right) \\ & 5\left|22^2+1,5\right| 23^2+1 \\ & \Rightarrow \quad 52 \mid f(22) \\ & n=20, f(20)=\frac{1}{2}(19 \times 21)\left(20^2+1\right)\left(21^2+1\right)\left(19^2+1\right) \end{aligned}$$$Not divisible by any prime square$\Rightarrow \quad n=20