CAT 2021 Slot 3 · Quant · Question 18

The first inequality looks awkward, but both sides are non-negative, so squaring is safe and turns it into something that factorises.

{\left( 1+mn \right)}^{2}<{\left( m+n \right)}^{2} gives 1+2mn+{m}^{2}{n}^{2}<{m}^{2}+2mn+{n}^{2}, and the 2mn cancels:

{m}^{2}{n}^{2}-{m}^{2}-{n}^{2}+1<0, which factorises as \left({m}^{2}-1 \right)\left({n}^{2}-1 \right)<0.

A product is negative only when the two factors have opposite signs, so one of {m}^{2},{n}^{2} is below 1 and the other above. For integers, below 1 means the value is 0.

So exactly one of m,n is 0. Take m=0: then \left( -1 \right)\left({n}^{2}-1 \right)<0 forces {n}^{2}>1, that is \left| n \right|\ge 2. The second inequality becomes \left| n \right|<5, so \left| n \right|\le 4.

That gives n=\pm 2,\pm 3,\pm 4: 6 pairs. By symmetry n=0 gives 6 more.

The pair \left( 0,0 \right) fails, since then the product is \left( -1 \right)\left( -1 \right)=1, not negative. The total is 6+6=12.