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Find infinitely many triples $(a,b,c)$ of positive integers such that $a,b,c$ are in arithmetic progression and such that $ab+1$, $bc+1$, and $ca+1$are perfect squares.

The solution is:

Consider the Pell's equation $x^{2}-3y^{2}=1$. If $(r,s)$ is a solution, then the triple $(a,b,c)=(2s-r,2s,2s+r)$ is in arithmetic progression and $(2s-r)(2s)+1 = (r-s)^{2}$, $(2s-r)(2s+r)+1 = s^{2}$, and $(2s)(2s+r)+1 = (r+s)^{2}$

Can someone explain to me how the author came up with this solution? How did he know to use $x^{2}-3y^{2}=1$?

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    $\begingroup$ Maybe the author knew the solution before the question :-) I am not completely joking: someone observed that the solutions of that Pell equation have additional properties. $\endgroup$ – Janko Bracic Jan 26 '15 at 14:50
  • $\begingroup$ This was an example from Putnam and Beyond. So it is meant to be do-able. $\endgroup$ – 1-___- Jan 26 '15 at 14:53
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    $\begingroup$ See also the analogous problem for $4$ integers here. $\endgroup$ – Dietrich Burde Jan 26 '15 at 21:06
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(Old answer revised.)

I. Method.

Here is a sketch of how one can find that Pell equation (and others) from first principles. Let,

$$\begin{aligned} ab+1\;& = (p_1x+p_2y)^2\\ ac+1\;& = (p_3x+p_4y)^2\\ bc+1\;& = (p_5x+p_6y)^2 \end{aligned}\tag1$$

Since we wish $a,b,c$ to be in arithmetic progression, assume,

$$a,\,b,\,c = -q_1x+q_2y,\;q_3y,\;q_1x+q_2y$$

for unknown integers $p_i,\,q_i$.Expand $(1)$ and collect powers of $x,y$,

$$\begin{aligned} r_1x^2+r_4xy+r_7y^2 \;& =1\\ r_2x^2+r_5xy+r_8y^2 \;& =1\\ r_3x^2+r_6xy+r_9y^2 \;& =1\\ \end{aligned}$$

where the $r_i$ are in terms of the $p_i,\,q_i$. The above is a clue that a Pell equation may be involved. Then solve the system,

$$\begin{aligned} & r_1 = r_2 = r_3=1\\ & r_4 = r_5 = r_6=0\\ & r_7 = r_8 = r_9\\ \end{aligned}$$

which is quite easy to do.

II. Solution.

We find,

$$\begin{aligned} ab+1\;& = \big(x-(m+n)y\big)^2\\ ac+1\;& = (ny)^2\\ bc+1\;& = \big(x+(m+n)y\big)^2 \end{aligned}$$

where $a,\,b,\,c = -x+my,\;2(m+n)y,\;x+my,\,$ and $x,y$ solve the more general Pell equation,

$$x^2-(m^2-n^2)y^2=1\tag2$$

Since $a,b,c$ are to be in arithmetic progression, let $m = 2(m+n)$, so $m,n = 2,-1$ and $x^2-3y^2=1$ pops out.

III. Others.

Using the same method,

$$\begin{aligned} ab-1\;& = (my)^2\\ ac-1\;& = \big(x+(m+n)y\big)^2\\ bc-1\;& = \big(x+(m-n)y\big)^2 \end{aligned}$$

where $a,\,b,\,c = x+ny,\;x-ny,\;2(x+my),\,$ and $x,y$ solve the similar Pell equation,

$$x^2-(m^2+n^2)y^2=1\tag3$$

IV. Higher Powers

Also by solving a system of equations, given,

$$x^2-17y^2 =\pm1\tag2$$

then,

$$(13x^2 + 12x y - 17y^2)^4 + (13x^2 - 12x y - 17y^2)^4 = (239x^4 - 14x^2 y^2 - 289y^4)^2+1$$

As you may notice, Pell equations are useful with other Diophantine equations where one term is set equal to 1.

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  • $\begingroup$ Can you also tell me how the author even knew to use Pell's Equation to begin with? $\endgroup$ – 1-___- Feb 6 '15 at 13:59
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    $\begingroup$ If you spend enough time with Diophantine equations, if one term involves $x_0=1$, it is usually a clue that a Pell equation may apply. $\endgroup$ – Tito Piezas III Feb 6 '15 at 14:05
  • $\begingroup$ @user2770287: I've updated my answer. It should be informative when dealing with similar equations. $\endgroup$ – Tito Piezas III Feb 7 '15 at 6:55
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I thought as this task to generalize and use for any numbers. It turned out that you can do without calculations. For the system of equations:

$$\left\{\begin{aligned}&ab+T=x^2\\&ac+T=y^2\\&bc+T=z^2\end{aligned}\right.$$

Enough to factor the following number:

$$bc=(y+c)^2-T$$

Using these numbers you can easily write the solution of this system of equations.

$$a=b-c-2y$$

$$b=b$$

$$c=c$$

$$x=b-c-y$$

$$y=y$$

$$z=y+c$$

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