Question 715444
Call the two consecutive integers x and x+1.  Their product x*(x+1) or {{{x^2 + x}}} is less than the square of the smaller.  The smaller is x, its square is {{{x^2}}} and {{{x^2 + x < x^2}}}<P>
Subtract {{{x^2}}} from both sides.  x<0.  Then the larger of the two integers x+1 < 1.<P>
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.<P>
If x=-10, then x+1 = -9.  Product:  90.  Smaller squared = {{{-10^2}}} = 100.  <P>
Let's try an x outside the solution, and as a result an x+1 outside the solution.<P>
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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