solution for equation

Question

For $a^2+b^2=c^2$ such that $a, b, c \in \mathbb{Z}$

Do we know whether the solution is finite or infinite for $a, b, c \in \mathbb{Z}$?

We know $a=3, b=4, c=5$ is one of the solutions.

http://en.wikipedia.org/wiki/Pythagorean_triple#Generating_a_triple — lab bhattacharjee, Dec 24 '12 at 15:25
I'm guessing that you mean "Do we know whether the number of solutions is finite or infinite?" rather than "Do we know whether the solution is finite or infinite?". — Michael Hardy, Dec 24 '12 at 15:48

score 5 · Accepted Answer · edited Dec 24 '12 at 07:23

5

Assuming $m,n$ be any two positive integers such that $m < n$, we have:

$$a = n^2 - m^2,\;\; b = 2mn,\;\;c = n^2 + m^2$$

And then $a^2+b^2=c^2$.

edited Dec 24 '12 at 07:23

Brian M. Scott

616,228

answered Dec 24 '12 at 07:20

Mikasa

67,374

1

Small comment: To produce all triples, we need to allow the $a$, $b$, $c$ above to be multiplied by an integer constant $k$. – André Nicolas Dec 24 '12 at 07:29
nice observation! – bsdshell Dec 24 '12 at 08:03

score 3 · Answer 2 · 2012-12-24T07:34:05.310

3

To comment on whether the solution is finite or infinite note that if $(a,b,c)$ satisfy $a^2 + b^2 = c^2$, then so does $(ka,kb,kc)$ where $k \in \mathbb{Z}$. Hence, either no integer solution exists or infinite integer solution exists.

You have already observed that $(3,4,5)$ satisfies $3^2 + 4^2 = 5^2$. Hence, infinite solutions exist.

Babak Sorouh has given the parameterization, which generates almost all possible solutions $a,b,c \in \mathbb{Z}^+$ (without scaling).

edited Dec 24 '12 at 07:34

answered Dec 24 '12 at 07:27

Thanks for noting an important comment in the first paragraph. Thanks @Andre. – Mikasa Dec 24 '12 at 07:31

P.K. · Answer 3 · 2012-12-24T17:00:39.050

0

You can use your reasoning and conclude that there are infinite right triangles that you could draw, but that is not a good proof.

It is already known that $3^2 + 4^2 = 5^2.$ You can use this statement itself to prove that there are infinitely many Pythagorean Triplets.

Let's prove that there are infinitely many Pythagorean Triplets in the form $(3k)^2 + (4k)^2 = (5k)^2,\ \, k \in \mathbb{N}$.

(NOTE that you can apply this to any Pythagorean Triplet; it's a property).

We have $(3k)^2 + (4k)^2 = (5k)^2$, or $9k^2 + 16k^2 = 25k^2$. Divide both sides by $k^2$ (it is not zero). We have $9 + 16 = 25 \iff 25 = 25$. We just have proved that the statement is true for any $k \in \mathbb{N}$ and there are infinite elements in $\mathbb{N}$.

REMARK The general proof is related.

edited Dec 24 '12 at 17:00

answered Dec 24 '12 at 16:07

P.K.

7,742

I'm sorry, but I don't see how FLT tells us that the sets of solutions to $a+b=c$ and $a^2+b^2=c^2$ are infinite. Could you explain this? – Dec 24 '12 at 16:33
Yes, according to the context of FMT. I have edited the post to avoid such further confusions. :) – P.K. Dec 24 '12 at 16:47
I understand that you could say these equations are related, but I don't think that Wikipedia page says anything about FLT implying that the sets of solutions for those equations are infinite (they state that they have an infinite number of solutions, but not that is an indirect or direct consequence of FLT.) – Dec 24 '12 at 16:59
Might as well remove the Fermat's part. :\ – P.K. Dec 24 '12 at 17:00

solution for equation

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