For a standard proof without using infinite descent : Assume n=4 is possible, then there exist integers a,b,c,d,e,f,g,h,m (whose gcd is 1) such that (ax+b)^2+(cx+d)^2+(ex+f)^2+(gx+h)^2=m^2(x^2+7)
Take x=1,3,5, we get (a+b)^2+(c+d)^2+(e+f)^2+(g+h)^2=8m^2\quad...(1) (3a+b)^2+(3c+d)^2+(3e+f)^2+(3g+h)^2=16m^2\quad...(2) (5a+b)^2+(5c+d)^2+(5e+f)^2+(5g+h)^2=32m^2\quad...(3) Now we use the following lemma: If 8|p^2+q^2+r^2+s^2, then p,q,r,s are even. The proof is trivial.
Using the lemma on (3) twice, we get that 4|5a+b,5c+d,5e+f,5g+h. Hence 4|a+b,c+d,e+f,g+h. Combining with (1), we get that m is even. Using the lemma on (2) twice, we get that 4|3a+b,3c+d,3e+f,3g+h. So a,c,e,g are even, which implies that a,b,c,d,e,f,g,h,m are all even, contradicting the assumption that their gcd is 1 .