# 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?

Jun 12, 2016

See explanation...

#### Explanation:

Define the sequence:

${a}_{0} = 0$

${a}_{1} = 1$

${a}_{n + 1} = 4 {a}_{n} - {a}_{n - 1}$

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:

${\left(a , b , c\right)}_{n} = \left({a}_{n} , 2 {a}_{n + 1} , {a}_{n + 2}\right)$ for $n = 1 , 2 , 3 , \ldots$

Then:

${a}_{n} + {a}_{n + 2} = {a}_{n} + \left(4 {a}_{n + 1} - {a}_{n}\right) = 4 {a}_{n + 1} = 2 \left(2 {a}_{n + 1}\right)$

So ${a}_{n} , 2 {a}_{n + 1} , {a}_{n + 2}$ are in arithmetic progression.

We find:

$\textcolor{b l u e}{{a}_{n} {b}_{n} + 1 = {\left({a}_{n + 1} - {a}_{n}\right)}^{2}}$

$\textcolor{b l u e}{{b}_{n} {c}_{n} + 1 = {\left({a}_{n + 2} - {a}_{n + 1}\right)}^{2}}$

$\textcolor{b l u e}{{c}_{n} {a}_{n} + 1 = {a}_{n + 1}^{2}}$

$\textcolor{w h i t e}{}$
Proof by induction

$\underline{\text{Base case}}$

$\textcolor{b l u e}{{a}_{1} {b}_{1} + 1} = 1 \cdot 8 + 1 = 9 = {\left(4 - 1\right)}^{2} = \textcolor{b l u e}{{\left({a}_{2} - {a}_{1}\right)}^{2}}$

$\underline{\text{Induction steps}}$

If:

${a}_{n} {b}_{n} + 1 = {\left({a}_{n + 1} - {a}_{n}\right)}^{2}$

This expands to:

$2 {a}_{n} {a}_{n + 1} + 1 = {a}_{n + 1}^{2} - 2 {a}_{n} {a}_{n + 1} + {a}_{n}^{2}$

So:

${a}_{n + 1}^{2} = 4 {a}_{n} {a}_{n + 1} - {a}_{n}^{2} + 1$

$= \left(4 {a}_{n + 1} - {a}_{n}\right) {a}_{n} + 1$

$= {a}_{n + 2} \cdot {a}_{n} + 1$

$= {c}_{n} {a}_{n} + 1$

So:

$\textcolor{b l u e}{{a}_{n} {b}_{n} + 1 = {\left({a}_{n + 1} - {a}_{n}\right)}^{2} \implies {c}_{n} {a}_{n} + 1 = {a}_{n + 1}^{2}}$

If:

$2 {a}_{n} {a}_{n + 1} + 1 = {\left({a}_{n + 1} - {a}_{n}\right)}^{2}$

Then we find:

${\left({a}_{n + 2} - {a}_{n + 1}\right)}^{2}$

$= {\left(3 {a}_{n + 1} - {a}_{n}\right)}^{2}$

$= 9 {a}_{n + 1}^{2} - 6 {a}_{n} {a}_{n + 1} + {a}_{n}^{2}$

$= \left(8 {a}_{n + 1}^{2} - 2 {a}_{n} {a}_{n + 1} + 1\right) + \left({a}_{n + 1}^{2} - 4 {a}_{n} {a}_{n + 1} + {a}_{n}^{2} - 1\right)$

$= \left(2 {a}_{n + 1} \left(4 {a}_{n + 1} - {a}_{n}\right) + 1\right) + \left({\left({a}_{n + 1} - {a}_{n}\right)}^{2} - \left(2 {a}_{n} {a}_{n + 1} + 1\right)\right)$

$= 2 {a}_{n + 1} {a}_{n + 2} + 1$

$= {b}_{n} {c}_{n} + 1$

That is:

$\textcolor{b l u e}{{a}_{n} {b}_{n} + 1 = {\left({a}_{n + 1} - {a}_{n}\right)}^{2} \implies {b}_{n} {c}_{n} + 1 = {\left({a}_{n + 2} - {a}_{n + 1}\right)}^{2}}$

Then:

$\textcolor{b l u e}{{a}_{n + 1} {b}_{n + 1} + 1} = {b}_{n} {c}_{n} + 1 = 2 {a}_{n + 1} {a}_{n + 2} + 1 = \textcolor{b l u e}{{\left({a}_{n + 2} - {a}_{n + 1}\right)}^{2}}$

$\textcolor{w h i t e}{}$
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:

${a}_{0} = 0$

${a}_{1} = 1$

${a}_{n + 1} = 6 {a}_{n} - {a}_{n - 1}$

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.