A question related to $f(n)=n^4+n^2+1$

Question

If $$f(n)=n^4+n^2+1$$ then we have to evaluate $$\frac{f\left(3\right)f\left(5\right)f\left(7\right)f\left(9\right)f\left(11\right)f\left(13\right)}{f\left(2\right)f\left(4\right)f\left(6\right)f\left(8\right)f\left(10\right)f\left(12\right)}$$ which, when run in Desmos, returns 61. But obviously we can't evaluate this with brute force. The farthest I have reached is that $$n^4+n^2+1=n^2(n+1)+\frac{n+1}{n+1}=(n^2+\frac{1}{n+1})(n+1)$$ Any hints?

Answer 1

We note that $n^4+n^2+1 = (n^4+2n^2+1)-n^2 = (n^2+1)^2-n^2 = (n^2+n+1)(n^2-n+1)$

Writing the value of $(n^2+n+1) = g(n)$, the value of $g(n-1) = n^2-n+1$

So we get eg $f(n)=g(n)g(n-1)$

And $\frac{f(3)f(5)f(7)...f(13)}{f(2)f(4)f(6)...f(12)}$

is equal to $\frac{g(2)g(3)g(4)...g(12)g(13)}{g(1)g(2)g(3)g(4)...g(12)} = \frac{g(13)}{g(1)}$

This evaluates to $(13*14+1)/(1*2+1)$ or $\frac{183}{3} = 61$

Answer 2

By using Sophie German Identity

$$f(n)=n^4+n^2+1=n^4+2n^2+1-n^2=(n^2+1)^2-n^2=(n^2-n+1)(n^2+n+1)$$

$$\frac{f\left(3\right)f\left(5\right)f\left(7\right)f\left(9\right)f\left(11\right)f\left(13\right)}{f\left(2\right)f\left(4\right)f\left(6\right)f\left(8\right)f\left(10\right)f\left(12\right)}=\frac{7\times 13\times 21 \times 31\times 43\times 57\times 73\times 91\times 111\times 133\times 157\times 183}{3\times 7\times 13 \times 21\times 31\times 43\times 57\times 73\times 91\times 111\times 133\times 157}=61$$