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Perfect Square To Divisibility By 56 (Posted on 2010-04-01) Difficulty: 4 of 5
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.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Some Thoughts A start | Comment 3 of 14 |
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


If N=a^2+a
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
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