Let us denote @=a and $=b then a+b=m and ab=n where m,n are +ve integers.

We have to prove a,b are integers[ma]+[mb] is a perfect square which involves two parts.

(i)a,b[ma]+[mb]=[a²+ab]+[b²+ab]

Now,Both a²+ab and b²+ab are integers only,so,it becomes,

[ma]+[mb]=a²+2ab+b²=(a+b)² which is a perfect square.

Now,let us prove the CONVERSE.

(ii)[a²+ab]+[b²+ab] is a perfect square

But a²+ab-1<[a²+ab]a²+ab and

b²+ab-1<[b²+ab]b²+ab

So,(a+b)²-2<[a²+ab]+[b²+ab](a+b)²

Since a+b=m is an integer,m² and m²-2 are both integers.

Now,we observe that if p,q are +ve distinct integers then p²-q² is always greater than 2.So,there lies no integer which is a perfect square between m² and m²-2.But for [a²+ab]+[b²+ab] to be a perfect square,the equality holds in the above relation.

i.e;[a²+ab]+[b²+ab]=(a+b)²

[a²+ab] and [b²+ab] are both integers.

a(a+b)a as (a+b)=m is an integer.Similarly,we have,b

Hence proved.