Pythagorean Triangles
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Pythagorean Triangles

A Pythagorean triangle (or Pythagorean triple) is a right-angle triangle containing sides with lengths x, y, and z, where $x, y, z, \in \mathbb{Z^+}$, that is, all of the side lengths of the triangle are positive integers. Hence, these values x, y, and z satisfy Pythagorean's theorem that $x^2 + y^2 = z^2$. The smallest Pythagorean triangle is the set 3, 4, 5. We note that the greatest integer is always going to by the hypotenuse, hence it is true that $3^2 + 4^2 = 5^2$. Hence, any positive integer solution to $x^2 + y^2 = z^2$ will be a set that is a Pythagorean triple.


We will first distinguish what is called a fundamental solution to $x^2 + y^2 = z^2$, that is, a solution where $(x, y) = 1$ ($x$ and $y$ must be relatively prime). Suppose $x$ and $y$ are not relatively prime, and that $(x, y) = d$. Hence it follows that d divides both $x$ and $y$. Hence $d$ must also divide $z$, and we obtain that $(x/d)^2 + (y/d)^2 = (z/d)^2$. Since $(x/d, y/d) = 1$, we know that the solution cannot be reduced further. For example, $(3, 4, 5)$ is a fundamental solution. However $(6, 8, 10)$ is not, since (6, 8) = 2, and in fact, we can divide each term in this Pythagorean triangle by $2$ to get $(3, 4, 5)$ again. In essence, a fundamental solution is unique in that the ratio a:b:c is unique and cannot be reduced by a factor d.

Lemma 1: If $a$, $b$, and $c$ is a fundamental solution to $x^2 + y^2 = z^2$, then exactly one of $a$ or $b$ is even.
  • Proof: Suppose that both $a$ and $b$ are even. Then both $a$ and $b$ are divisible by $2$, and hence $(a, b) \geq 2$, which implies that $a$, $b$, and $c$ is only a regular solution and not a fundamental solution.
  • Additionally, suppose that both $a$ and $b$ are odd. Then $a \equiv 1 \pmod 4$ and $b \equiv 1 \pmod 4$. Hence $a^2 + b^2 \equiv 2 \pmod 4$. But we note that $a^2 + b^2 = c^2$, hence $c^2 \equiv 2 \pmod 4$. This is impossible though, since $c^2$ can only be congruent to $0$ or $1$ modulo $4$. Hence both $a$ and $b$ cannot be odd. $\blacksquare$
  • Proof: We know that one of $a$ or $b$ is even. Let's say $a$ is even without loss of generality. Then $b$ is odd. Hence $a \equiv 0 \pmod 2$ and $b \equiv 1 \pmod 2$. Hence $c^2 = a^2 + b^2 \equiv 1 \pmod 2$, and hence c is odd. $\blacksquare$
Lemma 3: If $r^2 = st$ and $(s, t) = 1$ then both $s$ and $t$ are squares.

This will be a valuable proof in deriving an expression for all fundamental solutions to Pythagoras' theorem.

  • Proof: We know that s and t can be written in terms of their prime power decompositions, that is $s = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$ and $t = q_1^{f_1}q_2^{f_2}...q_r^{f_r}$ where all of the $p$s and $q$s are primes.
  • Since $s$ and $t$ are relatively prime, we know that $p_i ≠ p_j$ for each $i \neq j$. Hence if $r^2 = st$, then it follows that $r^2 = p_1^{e_1}p_2^{e_2}...p_k^{e_k}q_1^{f_1}q_2^{f_2}...q_r^{f_r}$. Since $r^2$ is a square, then all of the exponents must be even, that is $e_i \equiv f_j \equiv 0 \pmod 2$ for each$i$, $j$. Hence $s$ and $t$ must also be squares.
Lemma 4: Suppose that $a$, $b$, and $c$ is a fundamental solution to $x^2 + y^2 = z^2$ where $a$ is even and $b$ is odd. There exists positive integers $m$ and $n$ where $m > n$, $m$ and $n$ are relatively prime, and one of $m$ or $n$ is even such that $a = 2mn$, $b = m^2 - n^2$, and $c = m^2 + n^2$.
  • Proof: We stated that $a$ is even, hence $a = 2r$ for some integer $r$. Hence $a^2 = 4r^2$. We know that $a^2 = c^2 - b^2$, so it follows that $4r^2 = (c + b)(c - b)$. We know that $b$ is odd, and from corollary 2 we know that $c$ is also odd. Hence $(c + b)$ is even and $(c - b)$ is even. Let's say $c + b = 2s$, and $c - b = 2t$. It thus follows that $4r^2 = (2s)(2t) = 4st$, and hence $r^2 = st$. By lemma 3, we know that $s$ and $t$ must both be squares.
  • Now let's add the equations $c + b = 2s$, and $c - b = 2t$ together. We hence get $c = s + t$.
  • We now want to show that $(s, t) = 1$. Suppose that $d | s$ and $d | t$. Hence $d | b$ and $d | c$. But $(b, c) = 1$, hence $d = 1$, and $(s, t) = d = 1$.
  • From all of this, we gather from lemma 3 that $s = m^2$ and $t = n^2$. Hence we get that since $a^2 = 4st = 4m^2n^2$, then $a = 2mn$. We also get that $b = m^2 - n^2$ and $c = m^2 + n^2$.
  • Now we know that $m > n$ since $b = m^2 - n^2$.
  • Additionally, $(m, n) = 1$. Suppose that $d | m$ and $d | n$ for some integer $d$. Hence $d | a$ and $d | b$. But $(a, b) = 1$ since $a$, $b$, and $c$ are a fundamental solution, and hence, $(m, n) = 1$.
  • Lastly, we know that one of $m$ or $n$ is odd. Suppose that both are even. Then $c^2 = m^2 + n^2$ does not work as proven in lemma 2. Suppose that both are odd. This implies that $b$ is even and $c$ is even, which is incorrect as we defined $a$ to be even, so $b$ must be odd, and we know that $c$ is always odd in a fundamental solution.
Theorem 5: If $a = 2mn$, $b = m^2 - n^2$, and $c = m^2 + n^2$, and $a$, $b$, $c$ is a solution to $x^2 + y^2 = z^2$, where $m > n$, $m$ and $n$ are both positive and relatively prime, and one of $m$ or $n$ is odd, then $a$, $b$, $c$ is a fundamental solution.

We will now prove that given these conditions the solution $a$, $b$, $c$ is fundamental.

  • Proof: Algebraically we obtain that:
\begin{align} a^2 + b^2 = (2mn)^2 + (m^2 - n^2)^2 \\ a^2 + b^2 = 4m^2n^2 + m^4 - 2m^2n^2 + n^4 \\ a^2 + b^2 = m^4 + 2m^2n^2 + n^4 \\ a^2 + b^2 = (m^2 + n^2)^2 \end{align}
  • Since $a^2 + b^2 = c^2$, then $c^2 = (m^2 + n^2)^2$. Square rooting both sides we obtain that $c = m^2 + n^2$.
  • We will omit the rest of the proof at this time.
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