# How to prove that the roots of this equation are integers?

Let there be an equation $a^2 + 4ab + b^2 - 121 = 0$ where I want to prove that a,b are integers. Then I want to find whether there are integer values of $b$ for which $a$ is also an integer. Let us consider the case for $a$ $\rightarrow$ $a = \dfrac{-4b \pm \sqrt{16b^2 - 4(b^2-121)}}{2} = -2b \pm \sqrt{3b^2 +121}$ Thereby the problem reduces to showing that $3b^2 +121$ is a perfect square for certain integer values of $b$. This is what I cannot do.

Note: I know I can give examples and all for $3b^2 + 121$ to be a perfect square, but I'm looking for something with a little more substance. Thanks for the help.

• I'm a little unclear about the original problem. The way you have stated it, I can have non-integer solutions as well. Commented Feb 22, 2016 at 16:42
• Not all solutions of this equation are integral. Commented Feb 22, 2016 at 16:42
• @Wojowu Hmm. Yeah, I had my misgivings. Is there any way to show that there is a pair of $a, b$ for which the equation holds? I know that a,b = 5, 4. But I only found this by plugging in values. And so I have no way of showing that there is an integral solution. Commented Feb 22, 2016 at 16:45
• $12b^{2} + 4\times 11^{2}$ is a square if and only if $(3b^{2} + 11^{2}) = s^{2} \iff 3|(s-11)(s +11)$ and the smallest positive $s$ for which this happens is $s = 14$, which yields $b^{2} = 5^{2}$. Commented Feb 22, 2016 at 17:51
• To all; there are infinitely many solutions with both $a,b$ integers. Doubtless the original question in some book or contest asked to find at least some of these. Answer posted below. Commented Feb 22, 2016 at 18:00

EDIT: There are infinitely many integer solutions. They can all be found. In particular, if you have one solution $(a,b)$ you get a new solution with $$(a,b) \mapsto (-b, a + 4 b).$$ Going in the reverse direction, $$(a,b) \mapsto ( 4a + b, -a).$$ For solutions to $a^2 + 4ab + b^2 = 121,$ we will almost always need $ab < 0.$

Um. The generator for the (oriented) integer automorphism group of the quadratic form $a^2 + 4ab + b^2$ is $$A = \left( \begin{array}{rr} 0 & -1 \\ 1 & 4 \end{array} \right),$$ with inverse $$A^{-1} = \left( \begin{array}{rr} 4 & 1 \\ -1 & 0 \end{array} \right).$$ The degree two linear recurrences that follow come from applying Cayley-Hamilton to $A,$ in that $A^2 - 4A + I = 0.$

In the original variables, we can collect all solutions into Fibonacci type sequences, as in $a_{n+2} = 4 a_{n+1} - a_n$ and $b_{n+2} = 4 b_{n+1} - b_n$ $$\begin{array}{rrrrrrrrrrr} -4736 & -1269 & -340 & -91 & -24 & -5 & 4 & 21 & 80 & 299 & 1116 \\ 1269 & 340 & 91 & 24 & 5 & -4 & -21 & -80 & -299 & -1116 & -4165 \end{array}$$ We can do the same thing for the solutions where both $(a,b)$ are divisible by $11.$ Oh, not only can we switch the variables, we can always negate both. So $(-5,-4)$ gives us $(5,4)$ and $(4,5).$ $$\begin{array}{rrrrrrrrrr} -616 & -165 & -44 & -11 & 0 & 11 & 44 & 165 & 616 & 2299 \\ 2299 & 616 & 165 & 44 & 11 & 0 & -11 & -44 & -165 & -616 \end{array}$$

ORIGINAL: take the invertible change of variables $$a = u - 2 v, \;\; \; b = v,$$ so that $a + 2 b = u.$ You have $$u^2 - 3 v^2 = 121.$$ There are the imprimitive solutions where $u,v$ are solutions of $u^2 - 3 v^2 = 1$ then multiply both by $11.$ There are infinitely many of these, this is called the Pell equation.

There are also infinitely many imprimitive solutions to $u^2 - 3 v^2 = 121.$ This begins with $1 - 3 \cdot 4 = -11,$ and Brahmagupta's formula leads to $13^2 - 3 \cdot 4^2 = 11^2.$

In both cases, given values $(u,v),$ changing to $(2u+3v, u + 2 v)$ gives the same value of $u^2 - 3 v^2.$ You should check that, important.

A full accounting of all $(u,v)$ pairs is given by combining the $(2u+3v, u + 2 v)$ formula with the Conway topograph.

Here are some answers with the topograph , then two books that talk about it:

Another quadratic Diophantine equation: How do I proceed?

How to find solutions of $x^2-3y^2=-2$?

Generate solutions of Quadratic Diophantine Equation

Finding all solutions of the Pell-type equation $x^2-5y^2 = -4$

how to solve binary form $ax^2+bxy+cy^2=m$, for integer and rational $(x,y)$

Find all integer solutions for the equation $|5x^2 - y^2| = 4$

Maps of primitive vectors and Conway's river, has anyone built this in SAGE?

Infinitely many systems of $23$ consecutive integers

Finding integers of the form $3x^2 + xy - 5y^2$ where $x$ and $y$ are integers, using diagram via arithmetic progression

Small integral representation as $x^2-2y^2$ in Pell's equation

Solving the equation $x^2-7y^2=-3$ over integers

Solutions to Diophantine Equations

I'm afraid I was not able to fit this entire diagram onto one page. However, combine this with How to find solutions of $x^2-3y^2=-2$? and you get all integer expressions for $u^2 - 3 v^2 = 121,$ which then lead to all $a^2 + 4ab + b^2 = 121.$

All solutions $(u,v)$ come from the transformation $(2u+3v, u + 2 v),$ its inverse $(2u-3v, -u + 2 v),$ and the three triples $$(11,0), (13,4),(14,5)$$ Some care is need for the original $(a,b)$ problem because the change of variables does not quite keep positivity. Indeed, instead of the $\pm$ symmetries, the original variables have $(a,b)$ going to $(b,a).$

• I'm gonna be honest with you. You lost me after $u^2 - 3v^2 = 121$. As you've probably guessed, I have no experience in number theory. I am just a beginner in high school and I'm trying to learn the works of the trade. So if you could give some more layman explanations, that would be great. Commented Feb 22, 2016 at 18:31
• Furthermore, is there a way to find the solutions to this equation using no number theory at all? (Or at the most, very basic number theory?) Commented Feb 22, 2016 at 18:35
• I've edited my question to include a more basic, straightforward approach which I believe could work. Could you follow up on it, if you have the time? Commented Feb 22, 2016 at 18:44
• @S.Mo who gave you this question and why did they do that? What did they hope you would accomplish? Commented Feb 22, 2016 at 18:59

From $a^2+b^2+4ab-121=0$ we get $(a+b)^2+2ab=121$
Or $2=\frac{11+a+b}{a} \frac{11-a-b}{b}$
Now case 1: when the factors on right hand side are integers and coprime. Since 2 has only 2 factors, one of these has to be 1 and the other 2
Taking $\frac{11+a+b}{a}=2$ and $\frac{11-a-b}{b}=1$ we get $a=11 , b=0$
Similarly $\frac{11+a+b}{a}=1$ and $\frac{11-a-b}{b}=2$ gives $a=44$ and $b=-11$

I can't do the case where those factors are not integers or coprime right now (note that's the case of $a=5,b=4$)