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.
1. 56 = 7*8. The part of the solution showing 8 is a factor has already been done.<o:p></o:p>
2. For all m, m^2-1 = (m-1)(m+1)<o:p></o:p>
3. 3n+1 is a perfect square, x ^2. 3n= (x-1)(x+1) = k<o:p></o:p>
4. 4n+1 is a perfect square, y^2. 4n= (y-1)(y+1) = l<o:p></o:p>
5. 7n = k+l<o:p></o:p>
6. n=(k+l)/7<o:p></o:p>
7. n is an integer. Therefore n is divisible by 7.<o:p></o:p>
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Posted by broll
on 2010-04-04 04:09:42 |