Right-angled triangle with sides n²+1, n²-1, 2n

[liz]

the question:
(im presuming n^2 means n squared)

prove the following result: "a traingle with sides that can be written in the form n^2+1, n^2-1, and 2n (where n>1) is right angled.
show, by means of a counter example, that the converse is false.

this q was taken from the back of the book "the curious incedent of the dog in the night-time" and there was a full proof but i don't understand some of it.

it start by exmplaining we need to prove which side is the longest by doing:

n^2+1 - 2n = (n-1)^2

if n>1 then (n-1)^2 >0

therefore n^2+1 > 2n

similarily (n^2+1) - (n^2-1) = 2

therefore n^2+1 > n^2-1

so n^2+1 > n^2-1.

the rest is worked out using pythagoras but then the converse bit has completely lost me.

if anyone would like to explain, then i would be very grateful. thanks!

 

[Zurtex]

He is simply showing which side is longer,

$$(n^2+1) - (2n) = (n-1)^2$$

The left hand side is the difference between 2 of the sides.

However, $(n-1)^2 \geq 0$

So you can replace $(n-1)^2$ with: $\geq 0$ and you get:

$$(n^2+1) - (2n) \geq 0$$

Adding 2n to both sides:

$$n^2+1 \geq 2n$$

So you know that the n² + 1 side is greater or equal to the 2n side.

 

[dextercioby]

Please,do not double post.It's not really fair.

Daniel.