APMO 1989

We can suppose without loss of generality that $a,b,c,n\ge 0$. Let $(a,b,c,n)$ be a solution with minimum sum $a+b+c+n$. Suppose, for the sake of contradiction, that $a+b+c+n>0$. Since $6$ divides $5n^2$, $n$ is a multiple of $6$. Let $n=6n_0$. Then the equation reduces to $$6a^2+3b^2+c^2=30n_0^2.$$ The number $c$ is a multiple of $3$, so let $c=3c_0$. The equation now reduces to $$2a^2+b^2+3c_0^2=10n_0^2.$$ Now look at the equation modulo $8$: $$b^2+3c_0^2\equiv 2\left(n_0^2-a^2\right)(\mathrm{mod}8).$$ Integers $b$ and $c_0$ have the same parity. Either way, since $x^2$ is congruent to $0$ or $1$ modulo $4$, $b^2+3c_0^2$ is a multiple of $4$, so $n_0^2-a^2=(n_0-a)(n_0+a)$ is even, and therefore also a multiple of $4$, since $n_0-a$ and $n_0+a$ have the same parity. Hence $2(n_0^2-a^2)$ is a multiple of $8$, and $$b^2+3c_0^2\equiv 0(\mathrm{mod}8).$$ If $b$ and $c_0$ are both odd, $b^2+3c_0^2\equiv 4(\mathrm{mod}8)$, which is impossible. Then $b$ and $c_0$ are both even. Let $b=2b_0$ and $c_0=2c_1$, and we find $$a^2+2b_0^2+6c_1^2=5n_0^2.$$ Look at the last equation modulo $8$: $$a^2+3n_0^2\equiv 2\left(c_1^2-b_0^2\right)(\mathrm{mod}8).$$ A similar argument shows that $a$ and $n_0$ are both even. We have proven that $a,b,c,n$ are all even. Then, dividing the original equation by $4$ we find $$6\left(6(a/2{)}^2+3(b/2{)}^2+(c/2{)}^2\right)=5(n/2{)}^2,$$ and we find that $(a/2,b/2,c/2,n/2)$ is a new solution with smaller sum. This is a contradiction, and the only solution is $(a,b,c,n)=(0,0,0,0)$.