Suppose that 2M N +N define a interval of square numbers, 2N wide.
The distance between the extremes a^2 & b^2 can be formulate as n^2+2an, (n=ba), as they are squares numbers.
So:
2MN=a^2
2M+N=b^2
2N=b^2a^2=n^2+2an
And then:
P=(N/4)*(sq a^2/b^2)=((n^2+2a*n)/8)*
(a/a+n)=(n*a*(n+2a))/8(n+a)
In other words
P=(n*a/8)*[1+(a/(a+n))]
See that in the above expression:
[When ratio a/n decreases, the remainder of (a/a+n) is convergent to 0 and P > na/8 (but never reaching it).
And:
[When ratio n/a decreases, the remainder of (a/a+n) is convergent to 1 and P > na/4 (never reaching).
For each value of "a" there is no integer any more under a certain ratio in the P expression (and same for "n").
So for the higher P values a~n
With a=n
P=(a^2/8)*[(1+(a/2a)]=3a^2/16
Only fits for a=4 P=3 which is the same solution posted by Brian.
Edited on January 21, 2016, 1:43 am

Posted by armando
on 20160119 16:40:48 