Question

Here is a formula for generating the Pythagorean triples. If m and n are positive integers, with m < n, let a=n squared - m squared, b=2mn, and c=n squared + m squared. List the pythagorean triples that are generated using n < 4

Answer 1

If n < 4, and m<n, then the possible pairs of (m, n) are
(1, 2)
(1, 3)
(2, 3)
Now just compute the pythagorean triples using those values.

Answer 2

how did u get those number?

Answer 3

m, n are positive integers so \(0<m\) and \(0<n\) . Also, you know that \(m<n\) and \(n<4\). This implies that \(2\leq n\leq3\) and \(1\leq m\leq2\) . So then you find all pairs of numbers \((m,n)\) such that \(m \neq n\)

Answer 4

what so.. im confused lol..so is the answer (1,2) (1,3) (2,3)?

Answer 5

No. Those are the values of m and n that you use in your formula to get the pythagorean triples.

Answer 6

so i use the pythagorean theorem to find the answer?

Answer 7

You use the formula \[a=n^2-m^2\]\[b=2mn\]\[c=n^2+m^2\]with the three pairs of points I gave you to calculate the three different pythagorean triples.

Answer 8

OHH. ok. so there just those 3 points correct? or are there more?

Answer 9

Those are the only 3 possible points given the restrictions.

Answer 10

so for example. a=n^2-m^2
a=1^2-2^2
a=1-4
a=-3?

Answer 11

and then i plug it into b & c?

Answer 12

well to find the other 2 numbers

Answer 13

You switched the 1 and 2. It should be \(a=n^2-m^2=2^2-1^2=4-1=3\) so \(a=3\) As for finding b, and c, continue to use n=2, m=1.

Answer 14

I can do the first one as an example for you.

Answer 15

so its 3,4,5?

Answer 16

Let m=1, n=2. Then\[a=2^2-1^2=4-1-3 \Longrightarrow a=3\]\[b=2mn=2*1*2=4\Longrightarrow b=4\]\[c=n^2+m^2=2^2+1^2=4+1=5\Longrightarrow c=5\]So this gives the pythagorean triple \((3, 4, 5)\).

Answer 17

okay i get it. Also, how will i find an expression for m and n in terms of A,b,and c?

Answer 18

That's a little more tricky. I would suggest just adding a, b, and c together first and trying to solve from there.

Another idea would be to find \(a\cdot c\), and see where that gets you.

Answer 19

so you mean a+b+c=0? & i'll plug the things in later

Answer 20

\[a+b+c =(n^2-m^2)+2mn+(n^2+m^2)\]And try to isolate n or m. I would also recommend multiplying a, b, c together, and substituting either m or n into it.

Answer 21

ohh ok

Answer 22

I've got to go now, but feel free to tag me in a reply or another question. I'll look at it when I can.

Good luck!

Answer 23

ok thanks!