Natural Numbers Problem
Date: 12 May 1995 05:11:20 -0400 From: Dana Murray Subject: Question 5 Good morning Dr Math Thank you for the hints with question 4. How would you prove that one added to the product of four consecutive natural numbers is a perfect square? Thank you
Date: 16 May 1995 22:11:17 -0400 From: Dr. Sydney Subject: Re: Question 5 Hello again, Dana. Let's attack this problem by writing everything out in some helpful notation. For a general case, we can write the product of four consecutive natural numbers as n(n+1)(n+2)(n+3), where n is a natural number. If we were looking at the product 1*2*3*4, we would be looking at the n=1 case. If we were looking at the product 8*9*10*11, we would be looking at the n=8 case. Do you get the idea? So, our problem is to show that for any natural number, n, the number: n(n+1)(n+2)(n+3) + 1 is a perfect square. Again, let me give you a hint, and you can then see if you can take it from there, okay? Simplify the number, n(n+1)(n+2)(n+3) + 1, by multiplying through on the first term, combining like terms, etc. until you get a fourth degree polynomial. Can you factor this polynomial? The proof will be complete if you can factor the polynomial you get into a polynomial times itself. If you can't seem to see what your fourth degree polynomial would factor into, try this to figure it out. Write out a chart with 3 columns. Let the number n be in the first column, the number n(n+1)(n+2)(n+3) be in the second column (note don't multiply through, you'll be looking for a pattern here that is easier to see if you write this as a product of 4 numbers), and the number sqrt [n(n+1)(n+2)(n+3) + 1] (here sqrt means square root) be in the third column. Then the beginning of your chart will look like this (you can verify this by calculation...) 1 1*2*3*4 5 2 2*3*4*5 11 3 3*4*5*6 19 4 4*5*6*7 29 . . . . . . . . . Can you see any relations between the second and third columns? If you can, you come up with a relation between the first and second columns, then you won'tneed to factor that fourth degree polynomial. Do you see how? I'll let you work on this from here. If any of this seems confusing or unclear, write back, and I'll try and help. Good luck! Neat problem, too. --Sydney, "dr. math"
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum