1
$\begingroup$

There are $200$ balls in a pot numbered from $1$ to $200$.A ball is choosen at random.

What is the probability that Ball No. is a member of pythagorean triples?

Also find out the pythagorean triples.

Suppose I choose $31$,then how I find any two number $a,b$,so that,$31$ is member of pythagorean triples?

Is it manual or have some methods?

$\endgroup$
  • $\begingroup$ By "31 is a member of a Pythagorean triple" do you mean $a^2+b^2=31^2$, $31^2+a^2=b^2$, or both? $\endgroup$ – Semiclassical Jun 13 '16 at 4:52
  • $\begingroup$ damtp.cam.ac.uk/user/dv211/mathgaz03.pdf $\endgroup$ – lab bhattacharjee Jun 13 '16 at 4:53
  • $\begingroup$ a,b,c is member of pythagorean triples means any combination is possible,$a^2+b^2=c^2$ or $a^2+c^2=b^2$ or $b^2+c^2=a^2$.But greater than or less than is to be considered. $\endgroup$ – Hailey Jun 13 '16 at 4:54
  • $\begingroup$ Every odd number $2n-1$ is a member of a pythagorean triple: $(2n-1)^2 + (2n^2-2n)^2 = (2n^2-2n+1)^2$. It follows that every number is a member of a pythagorean triple, since you can just multiply all three members of that pythagorean triple by any constant you want, including any power of $2$. $\endgroup$ – Greg Martin Jun 13 '16 at 5:31
  • $\begingroup$ Greg martin. Every odd number except 1. So power of 2s might not be. As 3,4,5 is though, you can do all powers of 2 but 2. 1 and 2 can't be as $(n+1)^2 =n^2+2n+1>k^2+2$ for all $k\le n$. $\endgroup$ – fleablood Jun 13 '16 at 15:35
3
$\begingroup$

Notice $$\begin{align}(2k)^2 + (k^2-1)^2 &= (k^2+1)^2\\ (2k+1)^2 + (2k(k+1))^2 &= (2k^2+2k+1)^2 \end{align}$$ every integer $n \ge 3$ is part of a Pythragorean triple.

For the case $n = 31$, substitute $k = 15$ into $2^{nd}$ identity and you get $31^2 + 480^2 = 481^2$.

$\endgroup$
  • $\begingroup$ This is applicable for $n \ge 3$ ,that means probability is $\frac{198}{200}$ right? $\endgroup$ – Hailey Jun 13 '16 at 5:46
  • $\begingroup$ @Hailey, Yup. it should be $\frac{198}{200}$. $\endgroup$ – achille hui Jun 13 '16 at 5:48
0
$\begingroup$

If we include multiples like $(6,8,10)$, Pythagorean triples do indeed contain ever natural number $\ge 3$. However, if we include $only$ primitives, then the odd leg may be every odd number $\ge3$ but the even leg can only be a multiple of $4$. The odd leg will include $\frac{n}{2}-2$ and the even leg will include $\frac{n}{4}$ of all numbers from $1\text{ to }n$. Adding them together we get

$$\biggl(\frac{n}{2}-2\biggr)+\frac{n}{4}=\frac{(2n-8)+n}{4}=\frac{3n}{4}-2$$ So, with 200 balls we have $$\biggl(\frac{3*200}{4}-2\biggr)=150-2\text{ which gives a }\frac{148}{200}=74\%\text{ chance of drawing a Pythagorean triple number}.$$

For your question about $31$, you can find a triple for any side $A\ge3.$ We begin by solving $A=m^2-n^2$ for $n$ in terms of $m$.

$$A=m^2-n^2\Rightarrow n=\sqrt{m^2-A}\text{ where }\lceil\sqrt{A}\space\rceil\le m\le\frac{A+1}{2}$$

$$m_{min}=\lceil\sqrt{31}\space\rceil=6\qquad m_{max}=\frac{31+1}{2}=16$$ In our tests with $6\le m\le 16$, we find that only $m=6$ yields an integer $n=15$ and we have $$A=16^2-15^2=31\qquad B=2*16*15=480\qquad C=16^2+15^2=481$$

In a similar way, we can find out if there is a triple with a matching side $C$. $$\text{We let }n=\sqrt{C-m^2}\text{ where }1\le m\le \lfloor\sqrt{C}\rfloor$$

Whenever we get integer for $n$ and $n<m$ then we have the $(m,n)$ needed to find a Pythagorean triple with a hypotenuse equal to $C$. In the case of $31$, the range is $1$ to $5$. None of these values of $m$ yield an integer for $n$ so there is no triple with side $C=31$.

OTOH, if we pick $C=65$, we search $1\le m \lfloor\sqrt{65}\rfloor=8$ where $m=7$ and $m=8$ yield integers: $$f(7,4)=(33,56,65)\quad\quad f(8,1)=(63,16,65)$$ Then there is also $13X(3,4,5)=(39,52,65)$ and $5X(5,12,13)=(25,60,65)$

Now, we know there is also at least one triple where side $A=65$. Using the formula for $A$ above, searching $9\le m \le 33$, we find $$f(9,4)=(65,72,987)\quad\quad f(33,32)=(65,2112,2113)$$ And of course we have $5X(13,84,85)=(65,1092)$ and $13X(5,12,13)=(65,156,169)$

It seems we have $8$ distinct triples for the number $65$. This complicates things but it should make your investigation more interesting.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.