Each of X and Y is a positive integer, such that:
 X and Y are relatively prime, and:
 Each of (X^{2} – 5)/Y and (y^{2} – 5)/X is a positive integer.
Does there exist an infinite numbers of pairs (X,Y) satisfying the given conditions?
Give reasons for your answer.
..or,
more precisely, Lucas Aid, my memory having been
refreshed by your number sequence, Charlie.
1, 4, 11, 29, 76, …. , are alternate terms of the Lucas sequence {L_{i}}:
which looks like this: 1, 3, 4, 7, 11,
18, 29, 47, 76, ….
(i.e. Fibonaccilike but with a different starting pair.)
We know that L_{n} = L_{n1} + L_{n2} , from which it
can be deduced that
L_{n} = 3L_{n2} – L_{n4} which gives the rule for
Charlie’s sequence, {C_{i}}
as: C_{n} = 3C_{n1}
– C_{n2} (1) with C_{1} = 1 and C_{2} = 4
Noting that C_{n1}^{2} – 5 = C_{n}C_{n2} when
n = 3 (i.e. 4^{2} – 5 = 1*11), and
also that C_{n}^{2}
 5 = C_{n}C_{n2} + C_{n}^{2}
– C_{n1}^{2}
= C_{n}(3C_{n1} – C_{n})
+ C_{n}^{2} – C_{n1}^{2} using (1)
= C_{n1}(3C_{n} – C_{n1})
= C_{n+1}C_{n1}
we have proved by induction that C_{n1}^{2} – 5 = C_{n}C_{n2} is true for all
values of n >= 3. Since all the C values are integers (from (1)), it
follows that (C_{n1}^{2} – 5)/C_{n} and (C_{n}^{2}
– 5)/C_{n1} are always integers,
so every pair of consecutive C values satisfy the required condition.
Also, since C_{1} and C_{2} are coprime, it follows from (1)
that C_{2} and C_{3}
are coprime and, by induction, that all consecutive Cs are coprime.

Posted by Harry
on 20151231 17:52:16 