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 Divisible by 96? (Posted on 2016-06-05)
N is a positive integer such that:
each of N/2+1 and N/4+1 is a perfect square.

Is N always divisible by 96?
If so, prove it.
If not, provide a counterexample.

 No Solution Yet Submitted by K Sengupta No Rating

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 Another way Comment 2 of 2 |
N/2+1=P   N/4+1=Q   (P, Q, odd squares)

N=4(P-Q)=4(Q-1)   [so: P-Q=Q-1].

P-Q is a difference of odd squares which is always congruent to 0 (mod 8).

Q-1=q^2-1=(q+1)(q-1). This is congruent:
• to 0 (mod 3) if q is not multiple of 3.
• to 2 (mod 3) if q is multiple of 3.

a) If Q-1 is congruent to 0 (mod 3), then both (P-Q) and (Q-1) should be congruent with 0 (mod 24) (as P-Q=Q-1). Then N=4(P-Q) should be congruent to 0 (mod 96).

b) If Q-1 is congruent to 2 (mod 3) this is because Q=q^2 is and odd square multiple of 9 (q is multiple of 3: an odd one of course). Then it can be expressed as:  Q=(3(2n-1))^2

P is also an odd square. Then P=(2m-1)^2

P-Q=(2m-1)^2-(6n-3)^2=4(m(m-1)-9n(n-1)-2)
This expression is always congruent to 0 or 1 (mod 3). But Q-1 is congruent to 2 (mod 3). So it can't be P-Q=Q-1.Then no possible value for N.

Edited on June 7, 2016, 4:32 pm
 Posted by armando on 2016-06-06 08:40:40

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