My last post here is unconclusive. So I try again.
Suppose 2M -N +N define a interval of square numbers, 2N wide.
The extremes will be a^2 and b^2 and
the distance between the extremes will be n^2+2an, (n=b-a).
So: 2M-N=a^2 and 2M+N=b^2
P=(N/4)*(sq a^2/b^2)=((n^2+2a*n)/8)*(a/a+n) P=(n*a/8)*[1+(a/(a+n))]
This is a product of three factors, n, a, and [ ]/8. P should be a divisor of one of the three factors. But the third one is always between 2/8 and 1/8, so P is not a divisor here.
If P is divisor of a: a=kP, then kP*(n+2kP)/8(a+n))=1 an it results that P= n*(8-kn)/(2kkn-8) which is positive only for interval (4-kn-8) For K=1 n=6 a=3 and P=3
If P is a divisor of n: n=kP there is a quite similar formula with no positive value for P.
So the only solution is P=3 and no other solution is posible. (I edited again to add k factor)
Edited on March 1, 2016, 5:21 pm
Edited on March 2, 2016, 3:05 am
Posted by armando
on 2016-03-01 16:41:56