Pythagorean triples conditions 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)


*$x^2 + y^2 = z^2$ and

*$x + y > z$

*$x + z > y$

*$y + z > x$
(Because if 1., 2., 3. holds, we can construct right triangle with sides $x, y, z$.)
 A: 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?
A: One other way to analyse the equation $x^2+y^2=z^2$ is to solve it in its quadratic form.
Assuming $x\ne y$ otherwise $z$ would be irrational with factor $\sqrt{2}$.
Lets take $(x<y<z)>0$ without loss of generality and define $x=y-m$ and $z=y+n$ where $m$ and $n$ are positive integers. Giving
$$(y-m)^2+y^2=(y+n)^2\tag{1}$$
with $(z-x)=m+n$, $(z-y)=n$ and $(y-x)=m$
Which on rearrangement becomes
$$y^2-2(m+n)y+(m^2-n^2)=0\tag{2}$$
which can be solved as a quadratic in the usual way, with the result:
$$y=(m+n)+\sqrt{2n(m+n)}$$
[ or alternatively $y=(z-x)+\sqrt{2(z-y)(z-x)}$ ]
With a positive only square root since  $m$ and $n$ are always positive integers.
Therefore
$x=n+\sqrt{2n(m+n)}$, $y= (m+n)+\sqrt{2n(m+n)}$ and $z= (m+2n)+\sqrt{2n(m+n)}$
Then
$(1.)\;\;\;$ $x+y=(m+2n)+2\sqrt{2n(m+n)}>z\;(=(m+2n)+\sqrt{2n(m+n)})$,
$(2.)\;\;\;$ $y+z=(2m+3n)+2\sqrt{2n(m+n)}>x\;(=n+\sqrt{2n(m+n)})$
and
$(3.)\;\;\;$ $x+z=(m+3n)+2\sqrt{2n(m+n)}>y\;(=(m+n)+\sqrt{2n(m+n)})$.
(It is also true that $m\ge n$ or equivalently $3z \ge 5x$, but I'm not sure about the best way to prove that.)
(In the same manner $x^3+y^3=z^3$ can be converted to a cubic equation, $x^4+y^4=z^4$ to a quartic equation, $x^5+y^5=z^5$ to a quintic equation and so on)
Of course you don't need to go this far to prove statements (1.), (2.) and (3.).
Since it can be shown algebraically that $(x+y)^2$ is greater than $z^2$, this immediately implies $(x+y)>z$.
Since we assumed at the outset $z>y$, it trivially follows that $x+z>y$.
Since we assumed at the outset $z>x$, it trivially follows that $y+z>x$.
Recalling the Pythagorean solution $x = a^2-b^2, y = 2ab, z = a^2+b^2$, $y>x$ can't in general be assumed.
