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        riddles >> easy >> Diophantine squares
        
(Message started by: NickH on Apr 2^nd, 2003, 1:50pm)

Title: Diophantine squares
Post by NickH on Apr 2^nd, 2003, 1:50pm

Show that c² + 1 = (a² - 1)(b² - 1) has no solution in positive integers.

(Edited for towr's (and everyone else's) benefit!)

Title: Re: Diophantine squares
Post by towr on Apr 2^nd, 2003, 10:15pm

a = b = c = 0 => 1=1

now someone else can try with a,b,c > 0 :P

Title: Re: Diophantine squares
Post by wowbagger on Apr 3^rd, 2003, 2:55am

on 04/02/03 at 22:15:19, towr wrote:

now someone else can try with a,b,c > 0 :P

Let's suppose someone has already shown the nonexistence of solutions with a, b, c > 0. Based on this, I can rule out any solution with any one of a, b, c less than zero.
Does this mean we've already done more than half of the work? ;)

Title: Re: Diophantine squares
Post by LZJ on Apr 3^rd, 2003, 5:13am

But that's cheating!!! :o It's definitely not half the work, its no work at all.

Title: Re: Diophantine squares
Post by SWF on Apr 16^th, 2003, 7:26pm

c²+1=(a²-1)(b²-1)=(ab)²- a²-b²+1
c²=(ab)²-a²-b²

As long as a, b, and c are all even divide the equality above by 4 to give new values of a, b, and c satisfying the equality. Repeat until one or more of them is odd.

Odd integers squared have the form 4*n+1, while even integers squared have the form 4*n. With at least one of a, b, or c odd, it can be seen that the two sides of the equality above have a different remainder when divided by 4, so it is not possible for them to be equal. The c² has remainder 1 or 0. The other term has remainder 3 or 0, and only one side at a time can have remainder 0 since all the common factors of 4 have been removed.