# Pythagorean Triplets

The Pythagorean Theorem states that for a right triangle with sides a, b, and c, a2+b2=c2.

A Pythagorean triplet is a set of three numbers, where a2+b2=c2, a To extend this even further, a primitive Pythagorean triplet is a Pythagorean triplet where gcd(a,b,c)=1.

The goal is to find 100 primitive Pythagorean triplets.

Input: No input.

Output: 100 primitive Pythagorean triplets, in whatever order and shape, as long as the output contains only those triplets.

Shortest code wins. Good luck!

• Is a < b or are (a,b,c) and (b,c,a) 2 solutions. – user unknown Jun 22 '12 at 23:44
• @user unknown: a<b<c – beary605 Jun 23 '12 at 0:49
• They're called "triples", not "triplets". – mbomb007 Oct 28 '16 at 15:04

## Python, 42

i=4
while i<203:print 2*i,i*i-1,i*i+1;i+=2

• i**2 -> i*i to save a couple of characters. – breadbox Jun 23 '12 at 3:17
• I should mark this as cheating, but oh well. :) – beary605 Jun 23 '12 at 16:00
• How is it cheating? – grc Jun 24 '12 at 1:09
• "hoping for brute force" -- there's something you don't hear every day. – breadbox Jun 25 '12 at 16:14
• @beary605 why have you accepted the answer that isn't the shortest? – Griffin Jun 27 '12 at 15:35

# Javascript, 44 characters

Update:

for(i=s=0;i^400;)s+=[i+=4,r=i*i/4-1,r+2]+' '


I should have looked up an algorithm first. Well, can't beat python anyway.

Run in a console to see the result.

Old, slow, brute-force variant (140 characters)

for(q=a=[];563>++a;)for(b=a;--b;)(k=Math.sqrt(a*a+b*b))==~~k==function c(e,d,f){return f?c(c(f,e),d):d?c(d,e%d):e}(a,b,k)&&q.push([b,a,k]);q


## J, 646038 49 characters

100$}.~.(/:~*(0<*/*1=+./))"1[4$.$.(=/+/~)2^~i.699  Should've realised that such a big gain was probably flawed. Still uses the brute-force method, but done a bit more efficiently. Explanation: x=.2^~i.699 generates a list of ints from 0 to 699 and squares 2^~ them (~ here reverses the order of the arguments). (=+/~) is a hook that generates an addition table and compares the result to the list. This gives me a three dimensional array with items which are either 1 or 0. A 1 means that a2+b2=c2. $. converts to a sparse array. For my smaller (9) example I get:

   $.(=/+/~)2^~i.9 0 0 0 | 1 1 0 1 | 1 1 1 0 | 1 2 0 2 | 1 2 2 0 | 1 3 0 3 | 1 3 3 0 | 1 4 0 4 | 1 4 4 0 | 1 5 0 5 | 1 5 3 4 | 1 5 4 3 | 1 5 5 0 | 1 6 0 6 | 1 6 6 0 | 1 7 0 7 | 1 7 7 0 | 1 8 0 8 | 1 8 8 0 | 1  4$. just gives me the left part of this list.

(/:~*(0<*/*1=+./))"1[ decides which rows meet the criteria. (verb)"1 tells the verb to act on the individual rows rather than on the list. [ just separates the 1 and 4, otherwise J will think that 1 4 is a list of 2 numbers. '1=+./' gets the GCD of the three numbers and checks if it's 1. */ multiplies each triple together (getting 0 if it contains a 0) and these two are multiplied together. 0< turns the result into a boolean. * multiplies this result by each triple which eliminates all triples which contain a 0 or have a GCD which is not 1. /:~ sorts each triple.

~. selects the unique items from the list.

}. removes the first item (0 0 0) from the list.

100\$ takes the first 100 items from the list.

The above is my answer to this question, but as a matter of interest I implemented grc's method, (27 characters):

2(*,.<:@^~,.>:@^~)2*2+i.100


### Scala 180:

def g(a:Int,b:Int):Int=if(a==b)a else
if(a>b)g(a-b,b)else
g(b,a)
for(c<-(5 to 629);
a<-(1 to c-2);
b<-(a to c-1);
if((a*a+b*b==c*c)&&g(c,g(a,b))==1)&&a<400)println(a+"  "+b+" "+c)


Ungolfed:

def gcd (a: Int, b: Int) : Int = if (a == b)
a else
if (a > b) gcd (a-b, b) else
gcd (b, a)

def gcd (a: Int, b: Int, c:Int): Int = gcd (c, gcd (a, b))

def pythagorean (a: Int, b: Int, c: Int) = (a * a + b * b == c * c)

for (c <- (5 to 629);
a <- (1 to c-2);
b <- (a to c-1);
if (pythagorean (a, b, c) && gcd (a, b, c) == 1)) yield {
println (a + "² + "  + b + "² = " + c + "² => " + (a*a) + " + " + (b*b) + " = " + (c*c))
c*c
}


## APL (31)

↑{(2×⍵),(A-1),1+A←⍵×⍵}¨2+2×⍳100


It uses basically the same approach as the Python program.

# MATLAB 29

i=2:2:200;[2*i;i.*i-1;i.*i+1]


Mathematica 33

{2#,#^2-1,#^2+1}&/@Range[4,203,2]


# APL(NARS), 26 chars, 52 bytes

{k(4×⍵),2+k←¯1+×⍨2×⍵}¨⍳100


test:

  a←{k(4×⍵),2+k←¯1+×⍨2×⍵}¨⍳100
≢a
100
+/{(x y z)←⍵⋄(z*2)=(y*2)+x*2}¨a
100
+/{∨/⍵}¨a
100
10↑a
3 4 5  15 8 17  35 12 37  63 16 65  99 20 101  143 24 145  195 28 197  255 32 257  323 36 325  399 40 401


## R, 49 characters

translated from the Python solution:

i=4;while(i<203){print(c(2*i,i*i-1,i*i+1));i=i+2}

• You should take advantage of the functional nature of R i=seq(2,200,2);c(2*i,i*i-1,i*i+1) (though the output is a bit messy). – Griffin Jun 26 '12 at 13:55
• Ops! I posted without taking a look at all the other answers! :-P – Paolo Jun 26 '12 at 14:01