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(4A-x)     
C =  2B-A

   
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 = ((D-A)/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 2018-03-27 10:57:49