Seriously?! Let $a\ge b \ge c \ge d$ so that $4a^2\ge a^2+b^2+c^2+d^2=a^2b^2c^2d^2$. Then $4\ge b^2c^2d^2$ so $c=d=1$ and either $b=2$ or $b=1$. If $b=1$ then $a^2+3=a^2$ which I don't see happening and if $b=2$ then $a^2+6=4a^2$ i.e. $a^2=2$ which, again, I have difficulties believing.

xeroxia wrote: Show that the equation $a^2+b^2+c^2+d^2=a^2\cdot b^2\cdot c^2\cdot d^2$ has no solution in positive integers. Using modulo $4$, we can see that all of them must be even. So $a$, $b$,$c$, $d >1$ Hence, $a^2+b^2+c^2+d^2 <a^2b^2+b^2c^2<a^2b^2c^2d^2$

xeroxia wrote: Show that the equation $a^2+b^2+c^2+d^2=a^2\cdot b^2\cdot c^2\cdot d^2$ has no solution in positive integers. If all of them are odd,even=odd.So at least one of them is even,say $a$ Then $4|b^2+c^2+d^2$ If three of $b,c,d$ are odd,contradiction.If two odd and one even,then $b^2+c^2+d^2\equiv 2\mod 4,$ so not divisible by $4$ Then all of them is even.Now just repeat this process.