# [SOLVED] Right Triangle Problem

• Feb 7th 2009, 04:58 PM
KellHS886
[SOLVED] Right Triangle Problem
Prompt: List all integer possibilites for the three sides of a right triangle where one side is 2 units shorter than one of the other sides.

Is there some algebraic method of solving this?

Also, I found a 3-4-5 triangle and a 6-8-10 triangle. Both of these triangles have the same angles, which are 90 degrees, 36.86989765 degrees, and 53,13010235 degrees. Maybe this could mean something?

Any suggestions?
• Feb 7th 2009, 05:47 PM
oswaldo
From Pithagoras:
x^2 + (x+2)^2=y^2 where x and y anre ints.
You can investigate only-int solutions. May be a little complicated be should not be impossible.

OR (assuming the +2 side can be any side (hypo.) too)
x^2 + y^2 = (x+2)^2 gives
y^2 = 4X + 4
y = 2 * sqrt(x+1)

x= 8 gives y=6. All are integers.
x= 15 gives y=8, and x+2=17 all integers and fits the answer.
So, there should be infinitly many solution in the above format.
Please note Solution to the first expression may be hard but I suspect if there are infinitely many solutions from: x^2 + (x+2)^2=y^2 probably only 3, 4, and 5. But you got to check that out.

-O
• Feb 7th 2009, 08:07 PM
Soroban
Hello, KellHS886!

Quote:

List all integer possibilites for the three sides of a right triangle
where one side is 2 units shorter than one of the other sides.

Is there some algebraic method of solving this?

All "Pythagorean Triples" can be generated with these parametric equations:

. . $\begin{array}{ccc} a &=& m^2-n^2 \\ b &=& 2mn \\ c &=&m^2+n^2\end{array}\quad\text{ for positive integers }m\text{ and }n,\;m > n.$

It will be found that the required triples are generated for $n = 1$

. . $\begin{array}{|c|c||c|}\hline
m & n & (a,b,c) \\ \hline \hline
2 & 1 & (3,4,5) \\
3 & 1 & (8,6,10) \\
4 & 1 & (15,8,17) \\
5 & 1 & (24,10,26) \\
6 & 1 & (35,12,37) \\
7 & 1 & (48,14,50) \\
\vdots & \vdots & \vdots
\end{array}$