Exercise 5.3.16* (Triangles with integer sides and an angle of $60°$)

Question

Find, in the spirit of Theorem5.5, all primitive triples $x,y,z$ of positive integers such that a triangle with sides $x,y,z$ has an angle of $60°$.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} 
Consider a triangle with sides $x,y,z$, where  $x,y,z$ are positive integer, which has an angle of $60°$. Let $z$ is the length of the side intercepted by the angle $\gamma$ with measure 60°. Then the law of cosines gives
\begin{align*}
z^2 = x^2 + y^2 - 2xy \cos \gamma.
\end{align*}
Since $\gamma = \pi/3$, we obtain
\begin{align}
z^2 = x^2 + y^2 - xy.
\end{align}
Conversely, if $(x,y,z)$ satisfies (1), then $z^2 = x^2 + y^2 - xy = x^2 + y^2 - 2xy \cos \gamma$, thus $\cos \gamma = 1/2$, where $\gamma \in [0, \pi]$, so $\gamma = \pi/3$.

A triangle with sides $x,y,z$ has an angle of $60°$ if and only if some permutation of $(x,y,z)$ satisfies (1). It remains to find all primitive solutions $(x,y,z) \in (\N^*)^3$ of the equation (1), that is
\begin{align}
z^2 = x^2 + y^2 - xy,\qquad 	ext{ where } x \wedge y \wedge z = 1, \qquad x>0,\ y>0,\ z>0.
\end{align}
If $(x,y,z)$ satisfies (2), then $x,y$ cannot be both even , otherwise $z$ is even. This is impossible, because $ x \wedge y \wedge z = 1$. So $x$ or $y$ is odd. By exchanging $x$ and $y$ if necessary, we can assume that $y$ is odd. So we search the solutions $(x,y,z)$ of (2) with $y$ odd (which gives the solutions $(x,y,z)$ and $(y,x,z)$).

\bigskip
We write the equation (1) in an equivalent form:
\begin{align*}
z^2 = x^2 + y^2 - xy &\iff 4z^2 = 4x^2 + 4y^2 - 4xy\
&\iff (2z)^2 = (2x -y)^2 + 3 y^2\
&\iff 3y^2 = (2z - 2x +y)(2z + 2x -y).
\end{align*}

So we must find the solutions of 
\begin{align}
3y^2 = (2z - 2x +y)(2z + 2x -y),\qquad x \wedge y \wedge z = 1, \qquad (y 	ext{ odd}).
\end{align}

We show first that $a = 2z - 2x +y$ and $b = 2z + 2x -y$ are relatively prime. Suppose for the sake of contradiction that some prime $p$ divides these factors $a,b$, so that
\begin{align*}
p \mid 2z - 2x +y,\
p \mid 2z + 2x -y.
\end{align*}
Then $p \mid (2z - 2x +y) + (2z + 2x -y) = 4z$ and $p \mid (2z + 2x -y) - (2z - 2x +y) = 4x - 2y$.
We know that $p 
e 2$, otherwise  $p = 2 \mid y$, but $y$ is odd by hypothesis. Therefore $p \mid z$ and $p \mid 2x - y$. Moreover the equation (3) shows that $p^2 \mid 3 y^2$
\begin{itemize}
\item If $p = 3$, then $3^2 \mid 3 y^2$, thus $3 \mid y^2$, and $3$ is prime, so $3 \mid y$. Since $3 \mid 2x -y$, $3 \mid 2x$, so $3 \mid x$. Then $3 \mid x \wedge y \wedge z = 1$. This is a contradiction, which proves that $p
e 3$.
\item Since $p 
e 3$, $p^2 \mid 3 y^2$ implies $p^2 \mid y^2$. By the unique factorization theorem, $p \mid y$, and $p \mid 2x -y$, thus $p \mid 2x$ where $p
e 2$, so $p \mid x$. Then $p \mid x \wedge y \wedge z = 1$. This is a contradiction. 
\end{itemize}
Therefore $a \wedge b = 1$, where $3y^2 = ab$. Note that $a>0$ and $b>0$, otherwise $a\leq 0$ and $b \leq 0$, which implies $2z \leq 2x - u, \ 2z \leq -2x + y$, thus $4z \leq 0$, which is false. As in Problem 5, this shows that
$$(i)\ a = 3u^2,\ b = v^2\qquad 	ext{or} \qquad (ii)\  a= v^2,\ b = 3u^2,$$
where $u>0 0,v >0$ are suitable integers such that $u \wedge v = 1$. In both cases, $3y^2 = ab = 3u^2 v^2$, where $z>0, a > 0, b > 0$, thus $y = a b$. We examine these two cases.

\begin{enumerate}
\item[(i)] In the first case,
\begin{align*}
2z - 2x + y & = 3u^2,\
2z + 2x - y &= v^2.
\end{align*}
Then $4z = 3u^2 + v^2,\  -4x + 2y = 3u^2 - v^2$. Since $y = uv$ is odd, $u$ and $v$ are odd (and relatively prime). We obtain
\begin{align}
\left\{
\begin{array}{ll}
x &= \frac{2uv - 3 u^2 + v^2}{4},\
y &= uv,\
z & = \frac{3u^2 + v^2}{4}.
\end{array}
ight.
\end{align}
Since $u, v$ are odd, $u^2 \equiv v^2 \equiv 1 \pmod 4$, and $2uv \equiv 2 \pmod 4$, thus $3 u^2 + 4 v^2 \equiv 0 \pmod 4$ and $2uv - 3 u^2 + v^2 \equiv 0 \pmod 4$.

If we don't like quotients, we write $u = 2s + 1,\ v = 2t +1$, where $s \geq 0,t \geq 0$ are integers. If we replace $u,v$ in the preceding equations, we obtain
\begin{align}
\left\{
\begin{array}{ll}
x &= t^2 -3s^2 + 2st - 2s + 2t,\
y&= 4st + 2s + 2t +1,\
z &= 3s^2 + t^2 + 3s + t + 1.
\end{array}
 ight.
\end{align}
(we can take $s,t$ so that $(2s + 1) \wedge (2t + 1) = 1$).

Conversely, if (5) is true, then
\begin{align*}
x^2 + y^2 - xy =\  &      	
9 \, s^{4} + 6 \, s^{2} t^{2} + t^{4} + 18 \, s^{3} + 6 \, s^{2} t + 6\, s t^{2} \
&+ 2 \, t^{3} + 15 \, s^{2} + 6 \, s t + 3 \, t^{2} + 6 \, s +2 \, t + 1\
=\ & z^2.
\end{align*}
(Thanks to Sagemath!).
\item[(ii)] In the second case,
\begin{align*}
2z - 2x + y = v^2.\
2z + 2x - y = 3u^2.
\end{align*}
Then
\begin{align}
\left\{
\begin{array}{ll}
x &= \frac{2uv + 3 u^2 - v^2}{4},\
y &= uv,\
z & = \frac{3u^2 + v^2}{4}.
\end{array}
ight.
\end{align}
With $u = 2s + 1, v = 2t + 1$, this gives
\begin{align}
\left\{
\begin{array}{ll}
x &= -t^2 + 3s^2 + 2st +4s + 1,\
y&= 4st + 2s + 2t +1,\
z &= 3s^2 + t^2 + 3s + t + 1.
\end{array}
 ight.
\end{align}
Conversely, if (7) is true, then
\begin{align*}
x^2 + y^2 - xy =\  &      	
9 \, s^{4} + 6 \, s^{2} t^{2} + t^{4} + 18 \, s^{3} + 6 \, s^{2} t + 6\, s t^{2} \
& + 2 \, t^{3} + 15 \, s^{2} + 6 \, s t + 3 \, t^{2} + 6 \, s + 2 \, t + 1\
=\ & z^2.
\end{align*}
\end{enumerate}

In conclusion, if $(x,y,z)$ are positive integers such that $x \wedge y \wedge z = 1$, then
\begin{align*}
z^2 = x^2 + y^2 - xy  \iff &\exists u \in \N,\ \exists v \in \N,\  \exists \varepsilon \in \{-1,1\}, u\equiv v\equiv 1 \pmod 2 	ext{ and }\
&
\left\{
\begin{array}{ll}
x &= \frac{2uv  +\varepsilon( v^2 - 3 u^2)}{4},\
y &= uv,\
z & = \frac{3u^2 + v^2}{4}.
\end{array}
ight.
	ext{ or } 
\left\{
\begin{array}{ll}
x &= uv,\
y &= \frac{2uv  +\varepsilon( v^2 - 3 u^2)}{4},\
z & = \frac{3u^2 + v^2}{4}.
\end{array}
ight.
\end{align*}
(To avoid redundancy, we can take only the pairs $(u,v)$ such that $u \wedge v = 1$).

In another form,
\begin{align*}
z^2 = x^2 + y^2 - xy  \iff & \exists s \in \N,\  \exists t \in \N,\
& 
\left\{
\begin{array}{ll}
x &= t^2 -3s^2 + 2st - 2s + 2t,\
y&= 4st + 2s + 2t +1,\
z &= 3s^2 + t^2 + 3s + t + 1,
\end{array}
 ight.
 	ext{ or }
 \left\{
\begin{array}{ll}
x &= -t^2 + 3s^2 + 2st +4s + 1,\
y &= 4st + 2s + 2t +1,\
z &= 3s^2 + t^2 + 3s + t + 1.
\end{array}
 ight.\
\
 	ext{ or }
 & \left\{
\begin{array}{ll}
y &= t^2 -3s^2 + 2st - 2s + 2t,\
x&= 4st + 2s + 2t +1,\
z &= 3s^2 + t^2 + 3s + t + 1,
\end{array}
 ight.
 	ext{ or }
 \left\{
\begin{array}{ll}
y &= -t^2 + 3s^2 + 2st +4s + 1,\
x &= 4st + 2s + 2t +1,\
z &= 3s^2 + t^2 + 3s + t + 1.
\end{array}
 ight.
\end{align*}
To list the solutions, we only keep the triplets such that $x >0, y>0$.

(We can take only the pairs $(s,t)$ such that $(2s +1) \wedge (2t+1) = 1$).
\end{proof}

Note: Apart the obvious symmetry $(x,y,z) \leftarrow (y,x,z)$ which maps solutions on solutions, there is another symmetry, given by $(x,y,z) \leftarrow (y - x , y , z)$ (if $y>x$). Indeed
$$(y-x)^2 + y^2 - (y-x) y = x^2 + y^2 - xy = z^2.$$
We can explain this symmetry geometrically: if we complete a triangle with integer sides $x,y,z$ (where $y >x$) in an equilateral triangle, we obtain a second triangle with sides $(y-x, x, z)$ with an angle of 60°
(You can do a figure with the solution $(5, 8, 7)$ and the solution $(3, 8, 7)$).

The solutions of the second case are given by this symmetry: if 
$$x = \frac{2uv + 3 u^2 - v^2}{4},\  y = uv,\ z  = \frac{3u^2 + v^2}{4},$$
is a solution of the first case such that $y>x$, then
$$y-x = \frac{2uv - 3 u^2 + v^2}{4},\ y = uv,\ z  = \frac{3u^2 + v^2}{4},$$
is a solution for the second case.

\bigskip
Note: verification of these results with Python (on jupyter-lab)
\begin{verbatim}
from math import gcd, sqrt

def is_square(n):
    r = round(sqrt(n))
    if r * r == n:
        return True
    else:
        return False
        
l = []
for x in range(1,100):
    for y in range(1, 100):
        u = x ** 2 + y ** 2 - x * y
        if is_square(u):
            z = round(sqrt(u))
            if gcd(gcd(x,y), z) == 1 and y % 2 == 1: 
                l.append((x,y,z))
print(l)

[(1, 1, 1), (5, 21, 19), (7, 15, 13), (8, 3, 7), (8, 5, 7), (8, 15, 13), ...

l2 = []
for u in range(17):
    for v in range(18):
        for epsilon in [1,-1]:
            if u % 2 == 1 and v % 2 == 1  and gcd(u,v) == 1:
                 x, y, z = (2 * u * v + epsilon * (- 3 * u ** 2 + v ** 2)) // 4,
                                     u * v, (3 * u ** 2 + v ** 2) // 4
                if gcd(gcd(x,y), z) == 1 and x>0 and y>0:
                    l2.append((x,y,z))
print(l2)
[(1, 1, 1), (8, 5, 7), (15, 7, 13), (35, 11, 31), (48, 13, 43), (80, 17, 73),...
\end{verbatim}
We obtain the same solutions (but not in the same order). We have verified similarly (5) and (7).

Exercise 5.3.16* (Triangles with integer sides and an angle of $60°$)

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Exercise 5.3.16* (Triangles with integer sides and an angle of $60°$)

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