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
a(n)=4*a(n1)a(n2)
b(n)=2*a(n+1)
c(n)=a(n+2)
and write them in terms of a(n) and a(n1)
a(n)=a(n)
b(n)=8a(n)2a(n1)
c(n)=15a(n)4a(n1)
For a recursive proof assume the following to be perfect squares:
a(n)b(n)+1=8a(n)^22a(n)a(n1)+1
a(n)c(n)+1=15a(n)^24a(n)a(n1)+1
b(n)c(n)+1=120a(n)^262a(n)a(n1)+8a(n1)^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(n1))(30a(n)8a(n1))+1
=120a(n)^262a(n)a(n1)+8a(n1)^2+1
which is one of the ones we assumed above
Second product:
a(n+1)c(n+1)+1=(4a(n)a(n1))(56a(n)15a(n1))+1
=224a(n)^2116a(n)a(n1)+15a(n1)^2+1
this is not one of the above but is very close to c(n)^2
=225a(n)^2120a(n)a(n1)+16a(n1)^2
the difference being a(n)4a(n)a(n1)+a(n1)^21 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(n1))(56a(n)15a(n1))+1
=1680a(n)^2898a(n)a(n1)+120a(n1)^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 20180326 14:55:04 