(p²-q²)² = 1 + 16q
may be rewritten as:
[(p+q)(p-q)]² = 1 + 16q
p≠q otherwise the LS = 0 but RS cannot equal 0.
LS will always be positive because of the square format so q must also be positive for the RS to be positive. This however does not prevent p from being negative.
Since the LS is a square number, the RS must also be a square number, and it is odd.
If the absolute value of p and the value of q are different by one then either the p+q or the p-q value is 1 depending on whether p is +ve or -ve. This then reduces limits the rate at which the LS will increase. Howevr as the p value increases it causes the LS to outstrip the rate at which q generates the RS.
I have determined the following (p,q) values:
(4,5), (-4,5), (4,3) and (-4,3).
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Posted by brianjn
on 2007-09-01 23:42:28 |