N is a positive integer such that each of 3*N + 1 and 4*N + 1 is a perfect square.
Is N always divisible by 56?
If so, prove it. Otherwise, give a counterexample.
If N=3a^2+2a or N=3a^2-2a
then 3N+1 =
9a^2+6a+1=(3a+1)^2 or 9a^2-6a+1= (3a-1)^2 [the perfect squares for 3*N+1]
So N is of the form 3a^2+2a or 3a^2-2a
then 4N+1 =
4a^2+4a+1=(2a+1)^2 [perfect squares for 4*N+!]
So N is of the form a^2+a
What is sought is to prove that any numbers that are both of the first and second form are multiples of 56.
a^2+a = a(a+1) is always even. When are the others even? The even numbered terms are. Let a=2b
3(2b)^2+2(2b) = 12b^2+4b = 4(3b^2+b)
3(2b)^2-2(2b) = 12b^2-4b = 4(3b^2-b)
Which are actually multiples of 8. [I think I just have proven N must be a multiple of 8]
OK then when is a(a+1) a multiple of 8? Only when either a or a+1 is a multiple of 8. So it can be written as (8c)(8c+1) or (8c-1)(8c)
I just cant seem to get 7 to pop out of this...
Posted by Jer
on 2010-04-01 15:26:23