0
$\begingroup$

When you have one side of a right triangle fixed, is there a trend in pythagorean triples? For instance if I fix one side of a triangle at 1 unit what will the other side have to equal to get pythagorean triplets for the first 100 values? Is there a pattern to this?

$\endgroup$
  • 1
    $\begingroup$ Do you want integer side lengths? Rational ones? Aribitrary real ones? $(1,x,\sqrt{x^2+1})$ is not a very interesting generator of Pythagorean triples. $\endgroup$ – Henning Makholm Jun 8 '15 at 22:07
  • $\begingroup$ Maybe he wants $(1, \sqrt{x^2-1}, x)$? $\endgroup$ – Brian Tung Jun 8 '15 at 22:10
1
$\begingroup$

If one side (cathetus) is 1, then you cannot have a pythagorean triple because if $a$ is an integer, then $a^2 + 1$ cannot be a perfect square i.e. it cannot be equal to $b^2$ for some other integer b.

Pythagorean triple

Note: Usually by pythagorean triple is meant 3 positive integers, not just any 3 numbers.

$\endgroup$
  • $\begingroup$ So you cannot have a Pythagorean triple if one of the side lengths is 1? $\endgroup$ – Joe Jun 8 '15 at 22:13
  • $\begingroup$ Yes, that's correct (unless you mean non-integers are allowed too). $\endgroup$ – peter.petrov Jun 8 '15 at 22:14
1
$\begingroup$

Use rational numbers. 0.6 & 0.8 and 1.0. .36 + .64 = 1.0. The ancients used fractions, so 3/5 and 4/5.

$\endgroup$
1
$\begingroup$

You cannot do it with Pythagorean triples but you can do it with the sum of three cubes.

$$1^3+6^3+8^3=9^3=729$$ $$1^3+71^3+138=144=2985984$$ $$1^3+135^3+138^3=172^3=5088448$$ $$1^3+242^3+720^3=729^3=387420489$$ $$1^3+372^3+426^3=505^3=128787625$$ $$1^3+426^3+486^3=577^3=192100033$$ $$1^3+566^3+823^3=904^3=738763264$$

$\endgroup$
0
$\begingroup$

The identity $(2mn)^2 + (m^2-n^2)^2 = (m^2+n^2)^2$ is the basis of all Pythagorean triples. Normally, m,n are relatively prime positive integers with $m\gt n$. To get all triples, you may need to multiply each side by a fixed positive integer. If you allow rational sides, then

1 : $\frac{m^2-n^2}{2mn}$ : $\frac{m^2+n^2}{2mn}$

or

$\frac{2mn}{m^2-n^2}$ : 1 : $\frac{m^2+n^2}{m^2-n^2}$

could be used as triangle sides, and similarly you could make the hypotenuse equal 1. However, there are no Pythagorean triples with integer sides and one side equal to 1.

$\endgroup$
0
$\begingroup$

Is it OK if I use $2$ instead of $1$?

THEOREM


For all $a, b \in \mathbb Q^+$, $2^2 + a^2 = b^2$ if and only if there exists $\xi \in \mathbb Q^+$ such that $a = \xi - \dfrac{1}{\xi}$ and $b = \xi + \dfrac{1}{\xi}$.

PROOF
If $p, q, r \in \mathbb Z^+$ and $p^2 + q^2 = r^2$, then there exists $u, v \in \mathbb Z^+$ such that $(p, q, r) = (2uv, u^2-v^2, u^2+v^2)$. So there are two ways to parameterize a rational right triangle so that the length of one of the sides is 2.

\begin{equation*} \left(2,\; \dfrac{u^2-v^2}{uv},\;\dfrac{u^2+v^2}{uv}\right) \text{, or } \left( \dfrac{4uv}{u^2-v^2}, \; 2, \; \dfrac{2(u^2+v^2)}{u^2-v^2} \right) \end{equation*}

In the first case, we find that $\xi = \dfrac uv$. In the second case we find that $\xi = \dfrac{u+v}{u-v}$.

$\endgroup$
0
$\begingroup$

If you want sum of squares constant, then sides are

$(\cos(t), \sin(t),1) $

Instead, if you want difference of squares constant then

hypotenuse = $ \cosh(t) $, side = $ \sinh(t) $

The Pythagorean triplet is $ (\cosh(t), \sinh(t),1 ) $

You can choose $t$ interval and compute them.

Either way Pythagoras thm is obeyed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.