1x2x3x4 + 1 = 25 = 5^2 3x4x5x6 + 1 = 361 = 19^2 ... Prove that the product of 4 whole consecutive numbers plus one is always a perfect square. bye
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If by whole numbers you mean Z+, For any whole number n, n*(n+1)*(n+2)*(n+3)+1 = n^4+6*n^3+11*n^2+6*n+1 = (n^2+3*n+1)^2 (n^2+3*n+1)^2 is the square of (n^2+3*n+1), and (n^2+3*n+1) is an integer.
Last edited by paultwang; April 11th 2005 at 01:00 PM.
Originally Posted by paultwang If by whole numbers you mean Z+, As far as I can see (n^2+3n+1)^2 is a perfect square for all n \in Z, not only Z+...
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