# Problem on pythagorean triples

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?

• 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? – Semiclassical Jun 13 '16 at 4:52
• damtp.cam.ac.uk/user/dv211/mathgaz03.pdf – lab bhattacharjee Jun 13 '16 at 4:53
• 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. – Hailey Jun 13 '16 at 4:54
• 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 $2$. – Greg Martin Jun 13 '16 at 5:31
• 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 $k\le n$. – 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 $31^2 + 480^2 = 481^2$.

• This is applicable for $n \ge 3$ ,that means probability is $\frac{198}{200}$ right? – Hailey Jun 13 '16 at 5:46
• @Hailey, Yup. it should be $\frac{198}{200}$. – 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 $$n then we have the $$(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.