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?

  • $\begingroup$ 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

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 $31^2 + 480^2 = 481^2$.

  • $\begingroup$ 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

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.


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