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 Looking for n (Posted on 2004-02-24)
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.

 See The Solution Submitted by Aaron Rating: 4.2857 (7 votes)

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 Puzzle Solution Comment 13 of 13 |
(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.
Consequently,  it follows that the given expression cannot be a perfect cube.

Edited on September 5, 2022, 3:13 am
 Posted by K Sengupta on 2022-09-04 23:41:04

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