Question

Prove that every prime number is the leg of exactly one right triangle with integer sides.

~ Still not quite clear on how to prove this so any insight you have to offer will help! Thanks!

Answer 1

Draw a triangle with a base of 1, of 2, of 3, of....

So triangles with a leg for EVERY positive integer; I suppose you prove this by induction if you had to.

Answer 2

So...how would you go about proving that? I've noticed that in pythagorean triples there is normally atleast one prime number...so could I somehow use the pythagorean theorem to prove this?

Answer 3

It really depends on what u mean by a proof.

Are u arguing that I cannot draw such a series of triangles.

(Pythagorean triples won't help you with this one)

Answer 4

I really just am not understanding this problem...but I dont believe that in this case I could just use inductive reasoning for a proof. I found this one website
< http://mathforum.org/library/drmath/view/67289.html > that had an explanation but I'm still just not understanding

Answer 5

OK, you are going ahead of me with that, I just wanted to show you that you could construct a triangle with a leg equal to any integer to start with.

Answer 6

Let's instead ask what is it about the proof on mathforum that is confusing.

Answer 7

Okay~I'm understanding that to represent a prime number he uses p
and i understand that p^2 is odd but where is he getting f, g, and n, from? what numbers do f, g, and n represent?

Answer 8

'k, let me have a look...

Answer 9

The n^2 are just references to the series of squares and seeing that if you take one away from the next one, you get the prime "between"

Answer 10

The f and g follow from the previous statements to show by contradiction that there is only one triangle with a side length of p. ie they assume that there is another such triangle and show that it yields a contradiction.

Answer 11

Okay~ so basically he is saying that only one triangle can have a prime number side with that set of integers. That its not possible for that prime number to make a triangle with any other set of integers?

Answer 12

Essentially, yes.

The p comes from the differences in consecutive squares.

If you assume another triangle from non consecutive squares, then p is not square, a contradiction because you have already shown it to be square.

Answer 13

I worded that a bit badly, you see what I mean though?

Answer 14

Yes! I think I'm understanding this now!~ so
p^2 + n^2 =(n+1)^2
is proving that every prime number only has one set of integers to make it a right triangle
and that equation is basically derived, like you said, from the differences in the consecutive squares

Answer 15

Well, your equation above is the statement about consecutive squares yielding p squared.

THEN you go on to assume that this same p^2 can be found by some other difference of squares and what you find instead is that if this true, it contradicts with p^2 (it says p on the site) being square.

Answer 16

hmm...I'm afraid I lost you there..

Answer 17

OK, so 2 step process

1) These are the p^2

2) They are the only way to get the p^2

Answer 18

on your first step 1) what is these? what are the p^2?
(sry for having such a hard time comprehending this! :)

Answer 19

"These are the p^2" is what u wrote before, p^2 + n^2 =(n+1)^2

Answer 20

ie the p^2 come from the difference of consecutive squares.

Answer 21

Okay I understand that and I understand your step 2 (i think) that Its only through that equation you can get the difference of consecutive squares?

Answer 22

No, that's step 1.

Step 2 is the assumption7contradiction step.

Answer 23

Okay so step 1.) is basically the equation and that through that equation you get the difference of consecutive squares.
And step 2.) is where he plugs in the different integers (f and g) into the equation to contradict /prove that we can only get p^2 by using consecutive squares. So the squares used to derive that must be consecutive..(is that what he is trying to contradict?)

Answer 24

Yes, I think you have got it now, more or less.

You take the p^2 from Step 1 (which you have shown is square) and assume that f squared and g squared can generate it.

This f and g are not consecutive and unequal so cannot generate p squared.

Answer 25

Again I am wording this badly , the f and g squared can be factored into unequal terms so cannot generate p squared.

Answer 26

Okay I just re-read over the forum and I understand now what he's doing. So am I correct in saying that the process he used, he is proving that a prime number can be a leg for a right triangle with one set of consecutive squares but cannot be the leg of another triangle with different squares that are not consecutive. so therefore every prime number can be a leg of only one right triangle?

Answer 27

Right:-)

Answer 28

Hahaha! Well an hour later I got it! Thank you so much!! :) I appreciate it!! :)

Answer 29

Anytime, good workout for me as well...