Let n be the smallest positive integer such that n(n+1)(n+2)(n+3) can be expressed as either a perfect square or a perfect cube (not necessarily both).

Find n, or prove that this is not possible.

(I) The given expression can never be a perfect square.

(II) The given expression can never be a perfect cube.

EXPLANATION

(I) n(n+1)(n+2)(n+3)

= n(n+3)(n+2)(n+3)

=(n^2+3n) (n^2+3n+2.)

=P(P+2)

=(P+1)^2 -1

The given expression is precisely one less than a perfect square and, accordingly can NEVER correspond to a perfect square.

(II) Precisely one of n+1 and n+2 is odd,

(a) Let n+2 be odd.

Then, 2 must divide n and n+2.

Since each of n and n+2 is an odd number, this is not possible.

Also, n+2 and n+3 are relatively prime to each other,

Accordingly, for the expression to equal a perfect cube each of n+2 and:

n(n+1)(n+3) must be a perfect cube.

Now, we see that:

n(n+1)(n+3)=n^3+4n^2+3n

(n+1)^3=n^3+3n^2+3n+1

(n+2)^3=n^3+3n^2+3n+1

Then, we observe that:

(n+1)^3 < n(n+1)(n+3) < (n+2)^3, whenever n>1

Therefore, whenever n+2 is odd, the given expression van NEVER be a perfect cube, whenever n>1.

When n=1, the given expression assumes the numerical value of 24, which is not a perfect cube.

Consequently, the given expression fails to be a perfect cube whenever n+2 is odd.

(b) Let n+1 be odd.

Then, by arguments similar to the previous case, it follows that n, n+2, n+3 donor share a common factor.

Hence, each of (n+1) and n(n+2)(n+3) must be a perfect cube.

Now, we see that:

(n+1)^3 = n^3+3n^2+3n+1

n(n+2)(n+3)= n^3+5n^2+6n

Then, corresponding to integer values of n>0, it follows that:

(n+1)^3< n(n+2)(n+3) <(n+2)^3

Accordingly, n(n+2)(n+3) cannot be a perfect cube.

This leads to a contradiction.

Consequently, it follows that the given expression cannot be a perfect cube.

*Edited on ***September 5, 2022, 3:13 am**