jmc

Since the coefficients of $f(x)$ are all real, the nonreal roots come in conjugate pairs. Without loss of generality, assume that $a_1+i{b}_1$ and $a_2+i{b}_2$ are conjugates, and that $a_3+i{b}_3$ and $a_4+i{b}_4$ are conjugates, so $a_1=a_2,$ $b_1=-b_2,$ $a_3=a_4,$ and $b_3=-b_4.$

Then by Vieta's formulas, the product of the roots is $$\begin{array}{rl}(a_1+i{b}_1)(a_2+i{b}_2)(a_3+i{b}_3)(a_4+i{b}_4)& =(a_1+i{b}_1)(a_1-i{b}_1)(a_3+i{b}_3)(a_3-i{b}_3)\\ & =(a_1^2+b_1^2)(a_3^2+b_3^2)\\ & =65.\end{array}$$The only ways to write 65 as the product of two positive integers are $1\times 65$ and $5\times 13.$ If one of the factors $a_1^2+b_1^2$ or $a_3^2+b_3^2$ is equal to 1, then $f(x)$ must have a root of $\pm i.$ (Remember that none of the roots of $f(x)$ are real.) We can check that $\pm i$ cannot be roots, so 65 must split as $5\times 13.$

Wihtout loss of generality, assume that $a_1^2+b_1^2=5$ and $a_3^2+b_3^2=13.$ Hence, $\{|a_1|,|b_1|\}=\{1,2\}$ and $\{|a_3|,|b_3|\}=\{2,3\}$.

By Vieta's formulas, the sum of the roots is $$\begin{array}{rl}(a_1+i{b}_1)+(a_2+i{b}_2)+(a_3+i{b}_3)+(a_4+i{b}_4)& =(a_1+i{b}_1)+(a_1-i{b}_1)+(a_3+i{b}_3)+(a_3-i{b}_3)\\ & =2a_1+2a_3=6,\end{array}$$so $a_1+a_3=3.$ The only possibility is that $a_1=1$ and $a_3=2.$ Then $\{b_1,b_2\}=\{2,-2\}$ and $\{b_3,b_4\}=\{3,-3\},$ so the roots are $1+2i,$ $1-2i,$ $2+3i,$ and $2-3i.$ Then $$\begin{array}{rl}f(x)& =(x-1-2i)(x-1+2i)(x-2-3i)(x-2+3i)\\ & =[(x-1{)}^2+4][(x-2{)}^2+9]\\ & =x^4-6x^3+26x^2-46x+65.\end{array}$$Therefore, $p=\boxed{-46}.$