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.
(In reply to
More thoughts, problems encountered by Jer)
It seems a pity to leave a nice start unfinished. Here's how I approached it.
Start with the recurrence:
x = ((2 + sqrt(3))^n  (2  sqrt(3))^n)/(2 sqrt(3))
A = ((2 + sqrt(3))^(n+1)  (2  sqrt(3))^(n+1))/(2 sqrt(3))
B = 2(4Ax)
C = 2BA
Then with a little manipulation:
a = A
b = 2(sqrt(3A^2+1)+2A) = B
c = 4sqrt(3A^2+1)+7A = C
The arithmetic progression first, quite easy:
B  A 2(sqrt(3A^2+1)+2A)A = 2 sqrt(3A^2 + 1) + 3A = D [1]
C  B 4 sqrt(3A^2+1)+7A 2(sqrt(3A^2+1)+2A) = 2 sqrt(3A^2 + 1) + 3A = D, the same value.
But the next part looks messy to start with:
ab+1 2(sqrt(3A^2+1)+2A)*A+1 = 4A^2 + 2sqrt(3A^2 + 1)A + 1
bc+1 2(sqrt(3A^2+1)+2A)(4 sqrt(3A^2+1)+7A)+1 = 52A^2 + 30sqrt(3A^2 + 1)A+9 ac+1 A(4 sqrt(3A^2+1)+7A) = 7A^2 + 4sqrt(3 A^2 + 1)A
It's not at all obvious that these are squares.
But we can substitute in terms of A and D to simplify things, and the line of attack here is is to solve for D given what we know already about the squares from the examples given in earlier posts.
We can then solve to compute D in terms of A as in [1]:
ab+1 (D+A)*A+1 = ((DA)/2)^2; solving, D = 2sqrt(3A^2 + 1) + 3A
bc+1 (D+A)*(2D+A)+1 = 1/4(A+3D)^2; solving, D = 2sqrt(3A^2 + 1) + 3A
ac+1 (2D+A)*A+1 = 1/4(A+D)^2; solving, D = 2sqrt(3 A^2 + 1) + 3A
Since these solutions correspond with each other and the value in D in [1], we are done.
It's a nice problem, thanks Ady.
Edited on March 27, 2018, 11:05 am
Edited on March 27, 2018, 11:27 am

Posted by broll
on 20180327 10:57:49 