For $ n=1$: $ 4+4+4+4=\sqrt{4^{4}}$. For $ n=2$: $ 2+2+2+2=2\sqrt{16}$. For $ n=3$: $ 2+2+1+1=3\sqrt{4}$. For $ n=4$: $ 1+1+1+1=4\sqrt{1}$. Now, assume there was a solution $ a,b,c,d$ for $ n>4$, and take $ (a_{0},b_{0},c_{0},d_{0})$ the one where $ a+b+c+d$ is minimal, WLOG $ a_{0}\ge b_{0}\ge c_{0}\ge d_{0}$. Squaring both sides, it is natural to consider $ P(a)=a^{2}-a(n^{2}b_{0}c_{0}d_{0}+2(b_{0}+c_{0}+d_{0}))+(b_{0}+c_{0}+d_{0})^{2}$, the roots of which are integers $ a$ that are a solution of the problem. Since $ (b_{0}+c_{0}+d_{0})^{2}>0$, both roots $ a_{0}$ and $ a_{1}$ have the same sign, so both are solutions to the equation, and thus $ a_{1}\ge a_{0}\ge b_{0}\ge c_{0}\ge d_{0}$. So, $ 0\le P(b_{0})\leq (1+6+9)b_{0}^{2}-n^{2}b_{0}^{2}c_{0}d_{0}$, so $ n^{2}\le 16$, contradiction with $ n>4$.

Vieta jumping

The idea is correct but the writing is incorrect. Consider P(a) = x^2-n*sqrt(bcd)*x+(b+c+d). P(sqrt(b))>=0 leads to 4b>=n*b*sqrt(cd)>=n*b. Then n<=4.