Call the two consecutive integers x and x+1. Their product x*(x+1) or 
is less than the square of the smaller. The smaller is x, its square is 
and 
Subtract from both sides. x<0. Then the larger of the two integers x+1 < 1.
The problem has many solutions. For example if x=-1 then x+1 = 0 and the product 0*1 = 0 is less than the smaller squared = -1 squared = 1. 0 < 1.
If x=-10, then x+1 = -9. Product: 90. Smaller squared = 
= 100.
Let's try an x outside the solution, and as a result an x+1 outside the solution.
If x=0 then x+1 = 1. 0*1=0 and 0 squared = 0. The product is not less than the smaller number squared. That's what we expect since x+1 < 1 is the solution.
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