4
$\begingroup$

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.

  • 0
    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?".2012-12-24

3 Answers 3

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

  • 0
    nice observation!2012-12-24
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).

  • 0
    Thanks for noting an important comment in the first paragraph. Thanks @Andre.2012-12-24
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.

  • 0
    Might as well remove the Fermat's part. :\2012-12-24