A "Pythagorean Plus One" triple can be defined as any three distinct integers a, b, c, such all three of these are one more than a perfect square, and also a times b equals c.
What is the lowest value of c possible?
(In reply to
Maybe a really smart floobler..... by Penny)
Even a dumb one like me, perhaps! Check me on this:
If all three are even, then since a*b=c=z^2+1, z^2+1=4*k for some integer k. Hence z^2 must be congruent to 1 mod 4. Now z is either even (=2*n) so z^2=4*n^2 and is congruent to 0 mod 4, or z is odd (z=2*n+1) so z^2= 4*n^2+4*n+1 and is congruent to +1 mod 4. This is just a reiteration of the fairly wellknown fact that, modulo 4, a square must be congruent to 0 or 1.

Posted by Richard
on 20040517 14:54:34 