# Perfect Square relationship with no solutions

I would like to show that for positive integers $a>b,c$ all greater than 1 such that $c\nmid a$, there are no solutions to the following equation:

$$a^2+1=b^2(c^2+1)$$

As was pointed out in the comments, it might help to rearrange and factor:

$$(a-b)(a+b)=(bc+1)(bc-1)$$

Another interesting approach to the question, if we rewrite:

$$a^2-b^2c^2=b^2-1$$

Then we are trying to argue that $b^2-1$ cannot be squarefree (in particular, $c^2|b^2-1$)

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See @WillJagy's answer for an interesting twist... however, you could also rearrange the equation and factorize to see the properties: $(a-b)(a+b)=(bc-1)(bc+1)$ From this, you can extract which numbers need to have common factors for the equation to have solutions. –  orion May 9 '14 at 6:55

The conjecture is true. Fix $c$. The equation is equivalent to $$a^2 - (c^2+1) b^2 = -1$$ which is a "Pell equation", or in modern terminology the condition for $u = a + b \sqrt{c^2+1}$ to be a unit of norm $-1$ in ${\bf Z}[\sqrt{c^2+1}]$. By inspection we find a unit $\epsilon = c + \sqrt{c^2+1}$ of norm $-1$ (corresponding to the easy solution $(a,b)=(1,c)$). This unit is fundamental because the coefficient of $\sqrt{c^2+1}$ in $\epsilon$ is $1$, which is as small as a positive integer can get. Hence every unit has the form $u = \epsilon^n$ for some odd integer $n$, and the units with $b>0$ are those for which the exponent $n$ is positive.

In the binomial expansion of $\epsilon^n = (c + \sqrt{c^2+1})^n$ the terms that contribute to $a$ are ${n \choose 2m} c^{n-2m} (c^2+1)^m$ for $0 \leq m < n/2$. Each of those is clearly divisible by $c$, so $c\,|\,a$ as claimed.

This also confirms Will Jagy's belief that his "doubly infinite sequence" gives all solutions: his solutions indexed $0,1,2,3,\ldots$ correspond to $\epsilon, \epsilon^3, \epsilon^5, \epsilon^7$ etc.

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Exact same thing asked today; math.stackexchange.com/questions/788070/… –  Will Jagy May 9 '14 at 18:19

No. On the other hand, it appears that $c | a.$

I am watching a longer computer run, one infinite family turns out to be $$a = 4 c^3 + 3 c, \; \; \; b = 4 c^2 + 1$$ for any positive integer $c.$ You can check that $a^2 + 1 = b^2 (c^2 + 1).$

There are a few others, though, with especially small $c,$ which makes me think they might be derivable from the infinite sequence.

EEDDIITTTTT: that worked. I have a doubly infinite sequence, one infinite sequence for a given $c.$

$$a_0 = c, \; \; \; b_0 = 1,$$ $$a_1 = 4 c^3 + 3 c, \; \; \; b_1 = 4 c^2 + 1,$$ $$a_2 = 16 c^5 + 20 c^3 + 5 c , \; \; \; b_2 = 16 c^4 + 12 c^2 + 1,$$ then $$\color{magenta}{ a_{n+2} = (4 c^2 + 2) a_{n+1} - a_n, \; \; b_{n+2} = (4 c^2 + 2) b_{n+1} - b_n}.$$ I believe already that this doubly indexed family gives all solutions.

 a     b    c
38    17    2
117    37    3
268    65    4
515    101    5
682    305    2
882    145    6
1393    197    7
2072    257    8
2943    325    9
4030    401    10
4443    1405    3
5357    485    11
6948    577    12
8827    677    13
11018    785    14
a       b      c


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@ruadath, what is the source of this problem, and what is your background? –  Will Jagy May 9 '14 at 7:50

Unclear.

This equation is equivalent to the following : $a^2-qb^2=a^2-(c^2+1)b^2=-1$

You can use formulas solutions binary quadratic forms. It can be seen that the solutions of this equation is always there.

Though it is necessary to bring the decisions some pretty simple solutions:

the equation: $aX^2+bXY+cY^2=f$

If the root of the whole: $\sqrt{\frac{f}{a+b+c}}$

Then use the solution of Pell's equation: $p^2-(b^2-4ac)s^2=1$

Solutions can be written:

$Y=((4a+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$

$X=(-(4c+2b)ps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a+b+c}}$

If a root: $\sqrt{fa}$ then the solutions are of the form:

$Y=4ps\sqrt{fa}$

$X=(-2bps\pm(p^2+(b^2-4ac)s^2))\sqrt{\frac{f}{a}}$

Although it should be mentioned, and the equation: $aX^2-qY^2=f$

If the root of the whole: $\sqrt{\frac{f}{a-q}}$

Using equation Pell: $p^2-aqs^2=1$ solutions can be written:

$Y=(2aps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$

$X=(2qps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$

And for that decision have to find double formula.

$Y_2=Y+2as(qsY-pX)$

$X_2=X+2p(qsY-pX)$

So, take that formula and believe. Hopefully this will be enough.

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