Pythagorean triples conditions

algebra-precalculuselementary-number-theorypythagorean triples

Pythagorian triple is every triple of natural numbers $(x, y, z)$ such that $x, y, z$ are sides of a right triangle, where $z$ is the hypotenuse.

Now, Pythagorean theorem says:
$$x^2 + y^2 = z^2 \tag1$$

If we look just natural solutions to the equation $(1)$, without geometrical condition that a right triangle with sides $x, y, z$ do exist, do we have more solutions? And if yes, in which pattern?

Of course, condition that for natural triple satisfying $x^2 + y^2 = z^2$ there must exist right triangle with sides $x, y, z$, which is equivalent to add 3 conditions to the $(1)$: (triangle inequality)

  1. $x^2 + y^2 = z^2$ and
  2. $x + y > z$
  3. $x + z > y$
  4. $y + z > x$

(Because if 1., 2., 3. holds, we can construct right triangle with sides $x, y, z$.)

Best Answer

Here $a,$b and $c$ are the base, perpendicular and hypotenuse measures of the right angled triangle.

As per my understanding from OP's comments I need to prove that if $a^2+b^2=c^2$ then prove $a+b>c$, $b+c>a$ and $a+c>b$.

So $a^2+b^2 = c^2$

This implies that $(a+b)^2-2ab=c^2$

See $2ab$ will definitely be positive since $a$,$b$ and $c$ are natural number so adding just $2ab$ to lhs will make lhs $>$ rhs.

$(a+b)^2>c^2$

Rooting both sides

$(a+b)>c$

You might ask why didn't we take the negative case that's because a,b and c are natural numbers and sum of natural number is not negative.

Can you prove the other 2 cases yourself OP?

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