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 Infinite Illation Decision (Posted on 2015-12-31)
Each of X and Y is a positive integer, such that:
• X and Y are relatively prime, and:
• Each of (X2 – 5)/Y and (y2 – 5)/X is a positive integer.
Does there exist an infinite numbers of pairs (X,Y) satisfying the given conditions?

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes)

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 Lucozade (Spoiler) | Comment 2 of 3 |

..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 {Li}:
which looks like this:  1, 3, 4, 7, 11, 18, 29, 47, 76, ….
(i.e. Fibonacci-like but with a different starting pair.)

We know that Ln = Ln-1 + Ln-2 , from which it can be deduced that

Ln = 3Ln-2 – Ln-4 which gives the rule for Charlie’s sequence, {Ci}

as:        Cn = 3Cn-1 – Cn-2             (1)        with  C1 = 1 and C2 = 4

Noting that Cn-12 – 5 = CnCn-2 when n = 3 (i.e. 42 – 5 = 1*11), and

also that            Cn2 - 5 = CnCn-2 + Cn2 – Cn-12

= Cn(3Cn-1 – Cn) + Cn2 – Cn-12   using (1)

= Cn-1(3Cn – Cn-1)

= Cn+1Cn-1

we have proved by induction that   Cn-12 – 5 = CnCn-2  is true for all

values of n >= 3. Since all the C values are integers (from (1)), it

follows that (Cn-12 – 5)/Cn and (Cn2 – 5)/Cn-1 are always integers,

so every pair of consecutive C values satisfy the required condition.

Also, since C1 and C2 are coprime, it follows from (1) that C2 and C3

are coprime and, by induction, that all consecutive Cs are coprime.

 Posted by Harry on 2015-12-31 17:52:16

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