Suppose a and b are positive integers. We all know that aČ+2ab+bČ is a perfect square. Give an example where also aČ+ab+bČ is a perfect square. How many such examples exist?
For anyone interested, here is why (in one of the "base pairs") neither a nor b can be an even number but not divisible by 4 (2 mod 4).
The difference between two squares is always odd, or divisible by 4.
So, in a^2 + ab + b^2 = x^2, (a^2 + ab) is the difference between two squares, as is (b^2 + ab).
If a and b are both even, then of course they are not a "base pair" as they can be simplified to a smaller ratio.
So we are only concerned with the case when a is even and b is odd (or vice versa). Let us substitute for a, 4n-2, to show that it is 2 mod 4.
a^2 + ab = (4n-2)^2 + (4n-2)(b)
a^2 + ab = 16n^2 - 16n + 4 + 4nb - 2b.
All of the elements of the above equation are divisible by 4 until the last part. Since we know that b is odd, subtracting 2b makes the value of a^2 + ab 2 mod 4, which will not produce a square when added to b^2.
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Posted by tomarken
on 2006-04-26 15:34:41 |