Find Pythagorean Triple

Number Theory Level 5

For $a,b,c\in\mathbb N$ and $a,b<c$ there are $n$ Pythagorean Triple who satisfy the equation

$\large a+b+c=1000$

Find the value of $n$ and $a,b,c$ .

Submit your answer as $\large n+ \sum{abc}=??$

If $a,b,c\in\mathbb N$ and $(a,b,c)$ satisfy $a^2+b^2=c^2$ then $(a,b,c)$ is called Pythagorean Triple

The answer is 63750002.

2 solutions

Brian Charlesworth
Nov 26, 2017

We require that

$a + b + \sqrt{a^{2} + b^{2}} = 1000 \Longrightarrow \sqrt{a^{2} + b^{2}} = 1000 - (a + b) \Longrightarrow a^{2} + b^{2} = 1000^{2} + (a + b)^{2} - 2000(a + b)$

$\Longrightarrow a^{2} + b^{2} = 1000000 + a^{2} + b^{2} + 2ab - 2000a - 2000b \Longrightarrow ab - 1000a - 1000b = -500000$

$\Longrightarrow (a - 1000)(b - 1000) = 500000 \Longrightarrow (1000 - a)(1000 - b) = 500000 = 2^{5} \times 5^{6}$ .

Since we require that $0 \lt a,b \lt c \lt 1000$ , we will need to account for all (ordered) pairwise factorizations of $500000$ where both factors are less than $1000$ , in which cases we then assign one of the factors to $1000 - a$ and the other to $1000 - b$ . The only ways to achieve this is with $5^{4} \times (2^{5} \times 5^{2}) = 625 \times 800$ , or the reverse $800 \times 625$ , in which case, respectively, $1000 - a = 625 \Longrightarrow a = 375$ and $1000 - b = 800 \Longrightarrow b = 200$ , or the reverse $a = 200, b = 375$ . There are thus $n = 2$ possible triplets, namely $(a,b,c) = (375,200,425)$ or $(200,375,425)$ , and so

$n + \displaystyle\sum abc = 2 + 2 \times (200 \times 375 \times 425) = \boxed{63750002}$ .

After seeing the above solution. $We~ want~f(a,b)= a+b+c=1000. ~But~c=\sqrt{a^2+b^2},~~a~quadratic.\\ Also~f(a,b)=f(b,a), ~~\therefore~we ~only~have~two~solutions,~(a,b)~and~(b,a).~~n=2.\\ If~ x,y,z~ are~ primary~ triples~ such ~that~ for~an~ integer~ K,~a+b+c=K(x+y+z)=\Large K*S=1000.\\ Then~ x+y+z~ must~ be ~equal~ to~ a ~factor~ of ~1000.\\ Since~1000=2^3*5^2,~number~of~factors~are~(3+1)*(3+1)-2=14.\\ Factors~ of~ 1000=2,4,5,8,10,20,25,40,50,100,125,200,250,500.\\ So ~we~start~checking~primary~triples~and~reject ~triple~~with~S~\neq~any~of~the~factors~given~above.\\ ~~~~~~~\\ Triple~(3,4,5):-~~~~3+4+5=~12,~~reject~~~since~~\neq~any~of~the~above~factors. \\ Triple~(5,12,13):-~~~~~5+12+13=~30,~~reject, ~since~~~\neq~any~of~the~above~factors. \\ \color{#3D99F6}{Triple:-(8,15,17)~~~~8+15+17=S=~40,~~it~matches~with~factor~40.~so~K=1000/40=25.~\\ ~~~~~~~~~~~~~~~~~~~~~\therefore~~(a,b,c)=(25*8~ *~ 25*15 ~*~ 25,*17)=(200,375,425).~and~(375,200,425).}\\ So~n=2~and~answer~=\color{#D61F06}{2+~2*31875001=63750002.}.\\ If~the~triple~(a,b,c)~is~primary,~K=1.$

I missed it since I did not consider both (a,b,c) and (b,a,c).

Niranjan Khanderia - 3 years, 5 months ago

Md Mehedi Hasan
Nov 26, 2017

Find Pythagorean Triple

The answer is 63750002.

2 solutions

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