The Pythagorean Equation

OldTrout, Going Postal
Pythagoras had no time for tablets

We consider another quadratic Diophantine equation this time in three variables; namely, the Pythagorean Equation :

\(
x^2 + y^2 = z^2
\)

It is called the Pythagorean Equation although the Ancient Egyptians and Babylonians certainly knew how to generate Pythagorean triples.  Pythagorus used the formula \((2m, m^2 – 1, m^2 +1)\) which does not generate all the solutions.

A Pythagorean triple is an ordered triple \((x, y, z)\)  of positive integers which gives a solution to the above equation.  The ordered triple is said to be primitive if the greatest common divisor of \((x, y) = 1\).  The first such triple is, of course, \((3, 4, 5)\) since \(3^2 + 4^2 = 5^2 = 9 + 16 = 25\) and \(3\) and \(4\) are relatively prime.

A consequence of the condition that gcd \((x, y) = 1\) is that \(x\) and \(y\) cannot both be even.

Neither can \(x\) and \(y\) both be odd for the following reason:

All square numbers are congruent to either \(0\) (if even) or \(1\) (if odd) in modulo \(4\).

If \(x\) and \(y\) are both odd, then

\(x^2 + y^2\equiv 1 +1\equiv 2 \pmod 4 \)

and the sum of this addition is not a square number.

It follows that \(x\) and \(y\) have opposite parity and \(z\) will always be odd.

We chose to write the triples with the even value first \((4, 3, 5)\) and use the Euclidean formula for generating solutions.

\(x\equiv0\pmod 4\) , \(y\equiv1 \, \text{or} \, 3\pmod 4\, \text{and}\, z\equiv 1\pmod4 \) .

The primitive Pythagorean triples are generated as follows :

\((2mn, m^2 – n^2, m^2 + n^2)\),

where \(m\) and \(n\) are relatively prime positive integers of opposite parity and \(m\gt n\).

Observe that the even value, \(x = 2mn\), is always a multiple of \(4\).

All solutions of the equation are given by the Pythagorean triples \((x, y, z)\)

\(x = 2kmn,\: y = k(m^2 – n^2),\: z = k(m^2 + n^2)\),

where \(k\ge 1\) is any integer and \(m\) and \(n\) are relatively prime positive integers with opposite parity and \(m\gt n\).

The last just says that \((8, 6, 10)\) and \((12, 9, 15)\) . . . are scalar multiples of \((4, 3, 5)\).

The sequence of primitive triples begins

\(
\begin{array}{c c c}
m&n&\text{triple}\\
\hline
2&1&(4, 3, 5)\\
3&2&(12, 5, 13)\\
4&1&(8, 15, 17)\\
4&3&(24, 7, 25)\\
5&2&(20, 21, 29)\\
5&4&(40, 9, 41)\\
6&1&(12, 35, 37)\\
6&5&(60, 11, 61)\\
7&2&(28, 45, 53)\\
7&4&(56, 33, 65)\\
7&6&(84, 13, 85)
\end{array}
\)

 

We observed above that \(x\) is always a multiple of \(4\).

The product \(xy\) is always a multiple of \(12\) because \(x\) is a multiple of \(4\) and either \(x\) or \(y\) is a multiple of \(3\).

The product \(xyz\) is always a multiple of \(60\) with one of \(x, y\) or \(z\) being a multiple of \(5\).

 

\(\frac{(z – x)(z – y)}{2}\) is always a square number.

For example, using \((60, 11, 61)\)

\(\frac{(61 – 60)(61 – 11)}{2} = \frac{1\cdot 50}{2} = 25 = 5^2\).

 

There is an infinite family of Pythagorean triples where \(x\) is one less than \(z\).

 

\(
\begin{align}
\text{Let :} \qquad \qquad \qquad \qquad  m^2 + n^2 & = 2mn + 1\\
\text{Rearrange :} \qquad \quad \; m^2 – 2mn + n^2 & = 1\\
\text{Factorise :} \qquad \qquad \qquad  (m \, – n)^2 & = 1\\
\text{Take square roots :} \quad \quad \quad \; \; \; \,  m\, – n & = 1
\end{align}
\)

 

Hence,

\(
\begin{array}{c c c c c}
m&n&x&y&z\\
\hline
2&1&3&4&5\\
3&2&12&5&13\\
4&3&24&7&25\\
5&4&40&9&41\\
6&5&60&11&61
\end{array}
\)

 

There is an infinite family of near isosceles Pythagorean triples, that is when \(x\) and \(y\) differ by \(1\).

\(
\begin{align}
\text{Let : }\qquad \qquad \qquad \; x = 2mn\;\; \text{and}\; y & = m^2 – n^2\; \text{be consecutive integers}\\
\text{That is :} \qquad \qquad \qquad \qquad \; \; m^2 – n^2 & = 2mn \pm 1\\
\text{Rearrange :}\qquad \qquad \qquad \;\; m^2 – 2mn & = n^2 \pm 1\\
\text{Complete the square :}\quad m^2 – 2mn + n^2 & = n^2 + n^2 \pm 1\\
\text{Factorise :}\qquad \qquad \qquad \quad \; \, (m \, – n)^2 & = 2n^2 \pm 1\\
\text{Substitute :}\qquad \qquad \qquad \qquad \quad \; \; \, l & = m \, – n\\
\text{Rearrange :}\qquad \qquad \qquad \quad \; l^2 – 2n^2 & = \pm 1
\end{align}
\)

 

This is a Pell equation.  The solutions in \(l\) and \(n\) arise from the convergents of \(\sqrt 2\).

\(
\begin{array}{c c c c c c}
l&n&m&x&y&z\\
\hline
1&1&2&4&3&5\\
3&2&5&20&21&29\\
7&5&12&120&119&169\\
17&12&29&696&697&985\\
41&29&70&4060&4059&5741
\end{array}
\)

Or whenever \(\frac{m – n}{n}\) is a convergent of \(\sqrt 2\).

Primitive Pythagorean triples haves uses in cryptography.

 

© OldTrout 2017