Number of integer solutions of $a^2+b^2=10c^2$

Question

Find the number of integer solutions of the equation $a^2+b^2=10c^2$.

I can only get by inspection that $a=3m, b=m,c=m$ satisfies for any $m \in Z$.

Is there a formal logic to find all possible solutions? Any hint?

Also i tried taking $a=p^2-q^2$, $b=2pq$ and $10c^2=(p^2+q^2)^2$

which gives $$\frac{p^2+q^2}{c}=\sqrt{10}$$ which is invalid, since a rational can never be an irrational.

Answer 1

We first transform the given equation $a^2+b^2=10c^2$ as follows:

Consider $x=\frac{a}{c}, y=\frac{b}{c}$ (assuming we are interested in $c \neq 0$ case). Then we have to find rational points on the circle $$x^2+y^2=10.$$ Now $(x,y)=(3,1)$ is an obvious solution. To find other rational solutions, consider the line that passes through $(3,1)$ and has slope $m$. It is given by $$y=mx+(1-3m).$$ Consider the intersection of this line with the circle $x^2+y^2=10$. We can find the intersection from $$x^2+(mx+(1-3m))^2=10 \implies (1+m^2)x^2+2m(1-3m)x+(1-3m)^2-10=0.$$ But instead of solving the quadratic, we can argue that if two roots are $x_1$ and $x_2$, then we have $x_1=3$, so by means of Viete formula etc. we have $$x_1+x_2=-\frac{2m(1-3m)}{1+m^2} \implies x_2=\frac{3m^2-2m-3}{1+m^2}.$$ Now if $m$ is rational then $x_2$ is also rational and so will $y_2$ be, because $$y_2=\frac{1-m^2-6m}{1+m^2}.$$

Now for $m=\frac{p}{q}$, where $p,q \in \Bbb{Z}$ and $q \neq 0$, we can have $$\color{blue}{a=3p^2-2pq-3q^2, \quad b=q^2-p^2-6pq, \quad c=p^2+q^2}$$ Now the main question is: this combined with your answer (which can be obtained if we allow $q=0$) is that the totality of all solutions?

Answer 2

Let us consider the circle $C:x^2+y^2=10$. The question you asked is equivalent to finding all the rational points on this circle. Clearly, $(1,3)$ is one such point.

We will project from the point $(1,3)$ to the $Y$-axis. Let $(0,t)$ be the point of intersection of the line $L$ through $(1,3)$ and a point $(x,y)$ on the circle. Since $L$ passes through $(0,t)$ and $(1,3)$, its equation is \begin{align*} &x=\frac{y-t}{3-t}\\ \end{align*}

Now, the point $(x,y)$ is on the line $L$ as well as the circle $C$. So, \begin{equation*} \left (\frac{y-t}{3-t} \right )^2+y^2=10 \end{equation*} Solving this for $y$ and using $x=\frac{y-t}{3-t}$, we can get expressions for $x$ and $y$ in terms of $t$ only.

Now, since the $Y$-axis is a rational line, the rational points on circle must be mapped to rational points on the line. Also, from the geometry of this approach, it is clear that this gives us all the rational points (which means we get all the solutions of the equation you asked for).

Actually, for any given conic, if we know at least one rational point, we can get all others by this projection method. So, this is a very general approach. As you can see, in this case, we will get a very ugly expression for $t$- maybe a better choice of the initial rational point could have given better results. For a more detailed study, refer to chapter 1 section 1 of Rational Points on Elliptic Curves. They deal with the same problem, only with $x^2+y^2=1$ which gives a simpler expression of $t$.

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Edit: Wanted to add some links that may make reading this answer more interactive-

1. A Blog post that I wrote about the same problem
2. Slides of a talk I delivered on this problem and some generalizations

Answer 3

$$\color{magenta}{a=3p^2-2pq-3q^2, \quad b=-p^2-6pq+q^2, \quad c=p^2+q^2}$$

or, same thing

$$\color{green}{a=3x^2-2xy-3y^2, \quad b=-x^2-6xy+y^2, \quad c=x^2+y^2}$$

When $p,q$ are coprime, the primes that can still divide $\gcd(a,b,c)$ are $2$ and $5.$ The proof of all (primitive, integral) solutions is just showig that these don't matter.

When $x,y$ are both odd, all three of $a,b,c$ are divisible by $2,$ and we need to worry about whether half the triple is represented by the given parametrization. Well taking $$ p = \frac{x-y}{2} \; , \; \; q = \frac{x+y}{2} \; , \; \; $$

$$ 3 p^2 - 2pq -3q^2 = \frac{1}{2} \left( -x^2 - 6 xy + y^2 \right) $$ $$ - p^2 - 6pq +q^2 = \frac{-1}{2} \left( 3x^2 - 2 xy -3 y^2 \right) $$ $$ p^2 +q^2 = \frac{1}{2} \left( x^2 + y^2 \right) $$

$5$ is the other possibility . This happens when $$2x+y \equiv x - 2y \equiv 0 \pmod 5.$$

Taking $$ p = \frac{2x+y}{5} \; , \; \; q = \frac{x-2y}{5} \; , \; \; $$

$$ 3 p^2 - 2pq -3q^2 = \frac{1}{5} \left( 3x^2 - 2 xy -3 y^2 \right) $$ $$ - p^2 - 6pq +q^2 = \frac{1}{5} \left( -x^2 - 6 xy + y^2 \right) $$ $$ p^2 +q^2 = \frac{1}{5} \left( x^2 + y^2 \right) $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

The example I like to show is solving $$ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0, $$ four "recipes," $$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 37 u^2 + 51 uv + 8 v^2 \\ 8 u^2 -35 uv -6 v^2 \\ -6 u^2 + 23 uv + 37 v^2 \end{array} \right) $$

$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 32 u^2 + 61 uv + 18 v^2 \\ 18 u^2 -25 uv -11 v^2 \\ -11 u^2 + 3 uv + 32 v^2 \end{array} \right) $$

$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 38 u^2 + 45 uv + 4 v^2 \\ 4 u^2 -37 uv -3 v^2 \\ -3 u^2 + 31 uv + 38 v^2 \end{array} \right) $$

$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 29 u^2 + 63 uv + 22 v^2 \\ 22 u^2 -19 uv -12 v^2 \\ -12 u^2 -5 uv + 29 v^2 \end{array} \right) $$

For all four recipes, $$ x^2 + y^2 + z^2 = 1469 \left( u^2 + uv + v^2 \right)^2 $$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Answer 4

$$x^2+y^2=nz^2\tag{1}$$ In general, if equation $(1)$ has one integer solution $(x,y,z)=(x_0,y_0,z_0)$ then there exists an infinitely many integer solutions.

Substitute $x=t+x_0, y=t+y_0, z=at+z_0$ to equation $(1)$, then we get

$$t = \frac{2(x_0+y_0-nz_0a)}{-2+na^2}$$

Let $a=\frac{p}{q}$ with p,q are integers, hence we get a parametric solution $(x,y,z)=(x_0np^2-2qnz_0p+2q^2y_0, y_0np^2-2qnz_0p+2q^2x_0, nz_0p^2-2qx_0p-2qy_0p+2q^2z_0).$

Example of $x^2+y^2=10z^2.$

Let $n=10, (x_0,y_0,z_0)=(3,1,1).$

$(x,y,z)=(15p^2-10qp+q^2, 5p^2-10qp+3q^2, 5p^2-4qp+q^2).$

Answer 5

Solving the $C$-function of Euclid's formula $\quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$ for $(k), \space $ we can find Pythagorean triples for any given $C$-values, if they exist, that are primitive, doubles, or square multiples of primitives. This will not find, for example $(9,12,15)\space$ or $(15,20,25),$ but it will find $(3,4,5),\space (6,8,10),\space (12,16,20),\space (27,36,45), \space$ etc. We begin with the following formula. Any $m$-value that yields an integer $k$-value indicates a valid $(m,k)$ pair for generating a Pythagorean triple.

\begin{equation} C=m^2+k^2\implies k=\sqrt{C-m^2}\\ \text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor \end{equation} The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}.$

Here is an example for $C=40\implies 10c=4$ where $c$ is the one shown in the OP equation.

$$C=40\implies \bigg\lfloor\frac{ 1+\sqrt{80-1}}{2}\bigg\rfloor=4 \le m \le \lfloor\sqrt{40-1}\rfloor=6\\ \land \quad m\in\{6\}\Rightarrow k\in\{2\}\\$$ $$F(6,2)=(32,24,40)\implies (32,24,10\times 4)$$

This method will not find all Pythagorean triples that match the criteria but it will find an infinite number of triples that do such as:

$$c=1\longrightarrow (8,6,10\times 1)\\ c=2\longrightarrow (12,16,10\times 2)\\ c=4\longrightarrow (32,24,10\times 4)\\ c=9\longrightarrow (72,54,10\times 9)\\ $$

Note that any multiple of a triple found also yields a valid triple so $3\times (8,6,10)\longrightarrow (24,18,10\times3)$ and provides the "missing" $c=3$ triple in the list above. The combination of the two will find all Pythagorean triples where the $c$ in $10c$ is an integer except for the most unusual case like $(3,1,10\times 1)$ mentioned in another post.