How do you show that there are infinitely many triples #(a, b, c)# of integers such that #ab+1#, #bc+1# and #ca+1# are all perfect squares?

1 Answer
Jun 12, 2016

Answer:

See explanation...

Explanation:

Define the sequence:

#a_0 = 0#

#a_1 = 1#

#a_(n+1) = 4a_n-a_(n-1)#

This is A001353 in the online encyclopedia of integer sequences.

The first few terms are:

#0, 1, 4, 15, 56, 209, 780, 2911, 10864, 40545, 151316, 564719, 2107560, 7865521, 29354524, 109552575, 408855776, 1525870529, 5694626340, 21252634831, 79315912984, 296011017105, 1104728155436, 4122901604639, 15386878263120#

Then let:

#(a, b, c)_n = (a_n, 2a_(n+1), a_(n+2))# for #n = 1,2,3,...#

Then:

#a_n + a_(n+2) = a_n + (4a_(n+1)-a_n) = 4a_(n+1) = 2(2a_(n+1))#

So #a_n, 2a_(n+1), a_(n+2)# are in arithmetic progression.

We find:

#color(blue)(a_n b_n + 1 = (a_(n+1)-a_n)^2)#

#color(blue)(b_n c_n + 1 = (a_(n+2)-a_(n+1))^2)#

#color(blue)(c_n a_n + 1 = a_(n+1)^2)#

#color(white)()#
Proof by induction

#underline("Base case")#

#color(blue)(a_1 b_1 + 1) = 1*8 + 1 = 9 = (4-1)^2 = color(blue)((a_2 - a_1)^2)#

#underline("Induction steps")#

If:

#a_n b_n + 1 = (a_(n+1)-a_n)^2#

This expands to:

#2 a_n a_(n+1) + 1 = a_(n+1)^2-2 a_n a_(n+1) + a_n^2#

So:

#a_(n+1)^2 = 4a_n a_(n+1) - a_n^2 + 1#

#=(4 a_(n+1)-a_n)a_n + 1#

#=a_(n+2) * a_n + 1#

#=c_n a_n + 1#

So:

#color(blue)(a_n b_n + 1 = (a_(n+1)-a_n)^2 => c_n a_n + 1 = a_(n+1)^2)#

If:

#2 a_n a_(n+1) + 1 = (a_(n+1)-a_n)^2#

Then we find:

#(a_(n+2)-a_(n+1))^2#

#=(3a_(n+1)-a_n)^2#

#=9a_(n+1)^2-6a_n a_(n+1)+a_n^2#

#=(8a_(n+1)^2-2a_na_(n+1)+1)+(a_(n+1)^2-4a_na_(n+1)+a_n^2-1)#

#=(2a_(n+1)(4a_(n+1)-a_n)+1)+((a_(n+1)-a_n)^2-(2a_na_(n+1)+1))#

#=2a_(n+1)a_(n+2)+1#

#=b_n c_n+1#

That is:

#color(blue)(a_n b_n + 1 = (a_(n+1)-a_n)^2 => b_n c_n + 1 = (a_(n+2)-a_(n+1))^2)#

Then:

#color(blue)(a_(n+1) b_(n+1) + 1) = b_n c_n + 1 = 2 a_(n+1) a_(n+2) + 1 = color(blue)((a_(n+2)-a_(n+1))^2)#

#color(white)()#
Footnote

Iteratively defined sequences similar to the one for #a_n# above seem to be fairly common when finding sets of numbers with a particular property.

For example, the sequence of square roots of triangular square numbers is given by the recurrence relation:

#a_0 = 0#

#a_1 = 1#

#a_(n+1) = 6a_n - a_(n-1)#

The first few elements of this sequence are:

#0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175#

This is A001109 in the online encyclopedia of integer sequences.