maths plz help

a^2+b^2+c^2+d^2=3(a+b+c+d) 有多少組正整數解?
Sol
a^2+b^2+c^2+d^2=3(a+b+c+d)
(a^2－3a)+(b^2－3b)+(c^2－3c)+(d^2－3d)=0
(a^2－3a+9/4)+(b^2－3b+9/4)+(c^2－3c+9/4)+(d^2－3d+9/4)=9
(a－3/2)^2+(b－3/2)^2+(c－3/2)^2+(d－3/2)^2=9
(2a－3)^2+(2b－3)^2+(2c－3)^2+(2d－3)^2=36
36=1+1+9+25=9+9+9+9
(1) 36=1+1+9+25
2p－3=-1 =>p=1
2p－3=1 =>p=2
2p－3=3 =>p=3
2p－3=5 =>p=4
1，1，3，4=>4!/(2!1!1!)=12
1，2，3，4=>4!/(1!1!1!)=24
2，2，3，4=>4!/(2!1!1!1!)=12
(2) 36=9+9+9+9
2p-3=3 =>p=3
3，3，3，3=>4!/(1!1!1!1!)=1
12+24+12+1=49
有49組正整數解

abcd(a+b+c+d)=(a+b+c+d)/3
abcd=3