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.
Start with a couple of solutions as Dej Mar showed.
Then take the recurrance I noted before
and write them in terms of a(n) and a(n-1)
For a recursive proof assume the following to be perfect squares:
Now we just need to show each of the products +1
b(n+1)c(n+1)+1 are perfect squares
which is one of the ones we assumed above
this is not one of the above but is very close to c(n)^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
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