The product of two consecutive natural numbers is

Suppose $n^2+n$ can be a square. Then for some positive integer $m$ , $n^2+n=m^2$ . It is clear that $m>n$ . But $n=m^2-n^2=(m-n)(m+n)>m+n>n$ , which is a contradiction.

n and (n+1)are co-prime. ==> n(n+1) is a square iff n AND n+1 are squares. Which is never true!

Simple logic .. n(n+1) is never a square of a number

for the consecutive numbers,11,12,13,14,15------->the value of Y ,which is the product of n and n+1,is always even------>neccesarily may not be square.More over other options do not fit

If you need to multiply a number by itself to get a perfect square, then if you multiply a number by any different number it will not be a square

Don't forget those imaginary roots

“Y is always odd,” means the same thing as, “Y is never even,” so the only two possible choices are C and D. C can’t be right because n = 1 means the result is 2, which is prime. So D must be correct.

y=n^2+1 which is never a square....since discriminant is never 0