(2011-11-28) Product Of4 Sequential Whole Numbers

Number Two Son had a math problem positing that if you take 4 sequential whole numbers and multiple them together, then add 1, you always get a "perfect square" (number whose square root is a whole number). It played with a couple examples, but didn't prove the general assertion. So I got curious...

• generalize the total: [i*(i+1)*(i+2)*(i+3)]+1
• base example: i=1, product=24, total=25, root=5
• other examples seem to work, but can I prove it as general?
• expand the total to get i^4 + 6i^3 + 11i^2 + 6i + 1 - argh I can't remember how to factor polynomials like that! (Though it certainly looks promising with that pattern of factors...)
• hmm, that final root will be roughly the product of 2 of the terms. What if I take the greatest and the lowest together, then the 2 middle terms?
  • in our base example, I note that the products are 4 and 6, and the root is right between them! But what happens with bigger numbers?
  • i*(i+3) = i^2 + 3i
  • (i+1)*(i+2) = i^2 + 3i + 2
  • hey, the 2 products are always off by 2! So there will always be a number right between them. So, what if I take that number and square it?
  • [i^2 + 3i + 1]^2 = i^4 + 6i^3 + 11i^2 + 6i + 1 - booyah that matches the original total.

Boy I need a life.