How can I factor the polynomial $9x^4+16$ in $\mathbb{F}_{103}$ ?
My approach was take $x=\dfrac{2t}{\sqrt{3}}$, then $16t^4+16=16(t^4+1)=16((t^4+1)^2-2t^2)$ $=16(t^2-\sqrt{2}t+1)(t^2+\sqrt{2}t+1)$, but how can I factor this in $\mathbb{F}_{103}$? Thanks!
Because $103\equiv-1\pmod4$ we know that $-1$ is not a square in $K=\Bbb{F}_{103}$. Because $9$ and $16$ obviously are, we can conclude that the equation $9t^2=-16$ has no solutions in $K$. Therefore your polynomial cannot have a zero in $K$, so if it factors, it can only be a product of two quadratics.
So assume that $9x^4+16=a(x)b(x)$ for some pair of quadratic polynomials $a,b\in K[x]$. We can multiply here $a(x)$ by a constant as long as we at the same time divide $b(x)$ by the same constant. This allows us to adjust their leading coefficients to our liking. So without loss of generality we can assume that $$ a(x)=3x^2+a_1x+a_2,\qquad b(x)=3x^2+b_1x+b_2. $$ Expanding gives $$ a(x)b(x)=9x^4+[3a_1+3b_1]x^3+[3a_2+3b_2+a_1b_1]x^2+[a_1b_2+a_2b_1]x+a_2b_2. $$ Comparing the coefficients of degree three terms immediately gives $a_1=-b_1$. Using that bit of information in the linear term then implies that $a_2=b_2$ (excluding the possibility $a_1=0=b_1$ amounts to redoing the calculation in the first paragraph). The constant term then gives the constraint $a_2^2=16$. So we can conclude that $a_2=b_2=\pm4$.
Now we can rewrite the factors as $a(x)=3x^2+Cx\pm4$ and $b(x)=3x^2-Cx\pm4$ with the value of $C\in K$ and the sign of the constant terms still unknown. Expanding gives now $$ 9x^4+16=9x^4+[\pm24-C^2]x^2+16. $$ The choice of the unknown sign can be determined by determining which of the elements $\pm24$ is a square in $K$. The law of quadratic reciprocity gives that $2$ is a square (see DavidP's comment telling us that $2=38^2$), and that $3$ is a non-square. Therefore $24=2^3\cdot3$ is a non-square, but $-24$ is a square. So we need to use the minus sign and find $C\in K$ such that $C^2=-24$.
We easily see that $10^2=100=-3$, so $\sqrt{-3}=10$. Therefore $$ C=\sqrt{-24}=\sqrt{2^3\cdot(-3)}=2\sqrt{2}\cdot10=20\sqrt2=20\cdot38=39, $$ and the factorization is $$ 9x^4+16=(3x^2+39x-4)(3x^2-39x-4). $$