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Generating triplets of squares (Posted on 2018-03-25) Difficulty: 4 of 5
Find infinitely many triples (a,b,c) of distinct positive integers such that a, b, c are in arithmetic progression and ab+1,bc+1,ca+1 are perfect squares.

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts More thoughts, problems encountered | Comment 4 of 5 |
Start with a couple of solutions as Dej Mar showed.

Then take the recurrance I noted before
a(n)=4*a(n-1)-a(n-2) 
b(n)=2*a(n+1)
c(n)=a(n+2)
and write them in terms of a(n) and a(n-1)
a(n)=a(n)
b(n)=8a(n)-2a(n-1)
c(n)=15a(n)-4a(n-1)

For a recursive proof assume the following to be perfect squares:
a(n)b(n)+1=8a(n)^2-2a(n)a(n-1)+1
a(n)c(n)+1=15a(n)^2-4a(n)a(n-1)+1
b(n)c(n)+1=120a(n)^2-62a(n)a(n-1)+8a(n-1)^2+1

Now we just need to show each of the products +1
a(n+1)b(n+1)+1
a(n+1)c(n+1)+1
b(n+1)c(n+1)+1 are perfect squares

First product:
a(n+1)b(n+1)+1=(4a(n)-a(n-1))(30a(n)-8a(n-1))+1
=120a(n)^2-62a(n)a(n-1)+8a(n-1)^2+1
which is one of the ones we assumed above

Second product:
a(n+1)c(n+1)+1=(4a(n)-a(n-1))(56a(n)-15a(n-1))+1
=224a(n)^2-116a(n)a(n-1)+15a(n-1)^2+1
this is not one of the above but is very close to c(n)^2
=225a(n)^2-120a(n)a(n-1)+16a(n-1)^2

the difference being a(n)-4a(n)a(n-1)+a(n-1)^2-1 which equals zero using numbers of the sequence but I'm having trouble proving it!
So I'll leave that to later **edit:  I proved this recursively, but I don't feel like typing it in.**

The third product
b(n+1)c(n+1)=(30a(n)-8a(n-1))(56a(n)-15a(n-1))+1
=1680a(n)^2-898a(n)a(n-1)+120a(n-1)^2+1
this is not one of the above 
it appears to be a square for each value of a that I put in, but I cannot find a patern to try to prove.


Edited on March 26, 2018, 3:04 pm
  Posted by Jer on 2018-03-26 14:55:04

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