Right Triangle's Variable Sides

By Pythagoras, we have that

$(x + z)^{2} = (x + y)^{2} + (y + z)^{2}$

$\Longrightarrow x^{2} + 2xz + z^{2} = (x^{2} + 2xy + y^{2}) + (y^{2} + 2yz + z^{2})$

$\Longrightarrow 2xz = 2xy + 2yz + 2y^{2} = 2y(x + y + z) \Longrightarrow xz = y(x + y + z).$

Given that $y = z - 4,$ this last equation then becomes

$xz = (z - 4)(x + 2z - 4) \Longrightarrow xz = xz + 2z^{2} - 4z - 4x - 8z + 16$

$\Longrightarrow z^{2} - 6z - (2x - 8) = 0$

$\Longrightarrow z = \dfrac{6 \pm \sqrt{36 + 4(2x - 8)}}{2} = 3 \pm \sqrt{1 + 2x}.$

Now since $x$ and $z$ must be positive integers, we will require that $1 + 2x$ be a perfect square, and since $1 + 2x$ is odd this perfect square must be that of an odd number. (Also, as this odd perfect square will be $\ge 3$ we just need to look at the positive root, (since $z \gt 0$ ).) Thus for integers $k \gt 0$ we require that

$1 + 2x = (2k + 1)^{2} \Longrightarrow 1 + 2x = 4k^{2} + 4k + 1 \Longrightarrow x = 2k(k + 1).$

For $k = 1$ we have $(x,z) = (4,6),$ and for $k = 2$ we have $(x,z) = (12,8).$ In fact for $k \ge 2$ we will have $z \lt x,$ so since we must also satisfy the condition $z \gt x$ we are left with the unique solution $(x,y,z) = (4,2,6),$ which yields a $6/8/10$ right triangle and the ratio $\dfrac{z}{x} = \dfrac{6}{4} = \boxed{1.5}.$

Since x+z is only 4 greater than x+y, we can find any right triangle that satisfies that one of the leg is 4 less than the hypotenuse. Note that y+z needs to have length between the other two, so the right triangle has small integer length. 6-8-10 comes up as a matching triangle. 8 is between the other two, which means y+z=8. y+z=8 =>y+y+4=8 =>y=2 =>x=4 =>z=6 Hence 6/4 = 1.5.

Or do it the stupid long algebra way. That's also correct.

We have $y +4 = z$ , which is the same as $(x+y) + 4 = (x+z)$ . Therefore, the hypotenuse is $4$ more than one of the sides of the right angled triangle. Let $x+z = h+2$ and $x+y = h-2$ , for some $h> 2$ . Then by the Pythagoras' Theorem,
$(y+z)^2 = (h+2)^2 - (h-2)^2 = 8h$ , so $8h$ must be a perfect square(and it is even). Let $8h= 4k^2$ , for some natural number $k$ . Then $y+z = 2k$ , and $x+y = \frac {k^2}{2} - 2$ so $k$ is even and is greater than $2$ . Finally, $z> x$ implies $y+ z> y+ x$ , or $2k > \frac {k^2}{2} -2$ , or $k^2 - 4 k < 8$ . This means $(k-2)^2 < 12.$ Recall that $k$ must be even, so $k = 2, 4$ . But $k > 2$ . Therefore $k=4$ . So the sides of the triangle are $2 \times 4 =8, \frac {4^2}{2} - 2 = 6,$ and 10. Note that $2x+2y+2z = 6+8+10 = 24$ , so $x+y+z = 12$ Therefore $z = 12- (x+y) = 6 ,$ and $x= 12-(y+z) = 4$ . So $\boxed{\frac {z}{x} = 1.5}$

y= $\sqrt{1+2x}-1$

and z= $\sqrt{1+2x}+3$

x,y,z all are positive integers and ,

from triangle inequality we can say $z>x>y>0$ ,

z is also 4 more than 'y'

now,x=4 gives y=2 , z=6 therefore,

$\dfrac{z}{x}=1.5$

z=3 doesn't satisfy y+4=z but we have positive integers where they obey the above conditions

notice that i didn't work for the real triangle values for x,y,z

By Pythagoras, we have that

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

Substituting for y=z-4

(x+z)^2 = (x+z-4)^2 + (2z-4)^2

Expand, rearrange and factor yields

0 = (z-4)(z-2) - 2x

As x>0 ,(z-4)(z-2) must be >0 which implies that z<2 or z>4

Given that z>0, this implies that z=1 or z>4

As z>x, z=1 would imply that x (also a positive integer) would be 0, hence z cannot be = 1

Trialling z=5 (first integer greater than 4) yields a non-integer value for x

Trialling z=6 yields x=4 which satisfies all conditions and 6/4 = 1.5

$x+y+y+z \geq x+z$

$z \geq 4$

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

After much simplification

$2y^2+4y-4x=0$

Discriminant $=4^2-4 \cdot2\cdot 4x=16+32x$

Discriminant needs to be a perfect square as $x,y,z$ integers.

With some trial and error you get $x=4$

And after that $y=2$ and $z=6$ and $x=4$ .

So, $\dfrac{z}{x}=\dfrac{6}{4}=1.5$

By replacing y=z-4 and then using Pythagoras theorem We get z^2 - 6z +8 -2x=0 Now doing a bit of factorization we get (z-4) (z-2)=2*x

Now doing some logical comparison of both side we get z=6 ;x=4 So z÷x=1.5