## MMM #38: We have a winner!

Random.org selected Ken Lareau as the winner for MMM #38. Congratulations, Ken! Here's what the problem was:

Prove or disprove: The product of any four consecutive integers is always one less than a perfect square.

Don’t assume the integers are all positive.

Here's Ken's solution:

The above conjecture is true. The domain can be simplified a bit by noting two things:

1) If there are both positive and negative integers in the product, then since they're consecutive, 0 has to be present, meaning the product is 0, which is one less than 1, a perfect square.

2) If all the numbers are negative, their product is positive and equal to the product of the negation of each number (in other words, (-a) * (-b) * (-c) * (-d) = a * b * c * d).

Therefore we can restrict our validation to consecutive positive integers. From this, the proof is easy to see since if the first number is 'n', then we have for the product:

n * (n + 1) * (n + 2) * (n + 3)

which expanded is:

n^4 + 6n^3 + 11n^2 + 6n

One more than this is:

n^4 + 6n^3 + 11n^2 + 6n + 1

which happens to factor down to:

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

which is a perfect square. Hence the conjecture holds for all possible products.

I have one comment on the problems and on the solutions submitted. Many of the submissions ignored the part about not assuming all the integers to be positive. While it may be trivial to many of you that the signs of the numbers doesn't affect the solution, if you ignore that part of the problem then you're not solving the whole problem. Ken did a superb job of showing the signs don't matter. Yes, I realize that the algebraic solutions don't assume anything about the signs of the numbers but it's important to make that point in your argument.

And, here's a solution at Watchmath.com

Sue D. NymmeOctober 30th, 2009 - 13:53

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

The above formula is valid whether n is positive, negative, or zero. Why should anyone go to pains to describe a specific case when the general case holds?

SolOctober 30th, 2009 - 14:04

Sue,

The more I think about it the more I agree you’re right. In any case, I would have given the winner the benefit of the doubt if he or she had not said anything about the signs of the numbers.

Sue D. NymmeOctober 30th, 2009 - 14:14

Thanks, Sol. And congratulations to Ken!

watchmathNovember 3rd, 2009 - 18:10

Here is the link to my solution:

http://watchmath.com/vlog/?p=926