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New way to find Pythagorean Triples

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How to find all possible triples from only one side of right angle triangle

LOGICAL EXPLANATION

My formula is based upon the following equality

Hence if D is chosen in such a manner that it is a factor of x2 then we will always have integer values for  Pythagorean Triples since D will get canceled from numerator  &  denominator  Moreover D should be such that X2 / D is ‘Odd’ if D is Odd or it is ‘even’ if D is ‘even’ since even +/- even = even & odd +/- odd =even, denominator’s 2 will also get canceled. We should note that  D can also be greater than X, since we get negative value in such cases we should take absolute figure.

Thus if,

First side of triple = X then,

Second Side = |(X2 – D2)/2D|

Hypotenuses =  (X2 + D2)/2D

For defining solution set for D, we first have to express X in terms of 2* y. (where y is an odd number)

Let set ‘A’ contain factors of y2

Solution set for D =  {21 * all members of set A U 22 * all member of set A  U … 2n * all member of set A}

(Here ‘U’ stands for union)

 [Note:- While solving last part (i.e. 2n * all member of set A), one should stop as soon as 2n * any member of A = X and remaining members of A need not be consider since they will give same answer as first half element of that particular subset.]

Examples:

1) if X = 35 then it means X = 20 * 35 hence  y = 35.

Hence set A will contain factors of 352 i.e. {1,5,7,25,35,49,175,245,1225}


However Solution set for D will be {1,5,7,25} [as 20*5th element of set = 35 we need to consider only till 4th element]. Hence there will be 4 triples having 35 as one of the sides.

Triples in this case will be (35,612,613),(35,120,125),(35,84,91),(35,12,37)

2) if X = 24 then it means X =  23 * 3 hence y = 3.

3) if X=64 then it means X = 26 * 1  hence y=1 

Hence set A will contain factors of 12 i.e. {1}


Solution set for D will be {2x (1) U 2x (1) U 2x (1) U 24 x (1) U 2x (1) U 2x (1)}

i.e. {2,4,8,16,32} [since in last iteration, 26*1 = 64 we need to consider only till earlier element]. Hence there will be 5 triples having 64 as one of the sides.

4) if X=11 then it means X = 20 * 11  hence y=11

Hence set A will contain factors of 112 i.e. {1,11,121}

Solution set for D will be {1} [as 20*2nd element of set = 11 we need to consider only till 1st element]. Hence there will be 1 triple having 11 as one of the sides.

The only triple in this case will be (11,60,61). Thus we can note that when X is prime there will be only one pythagorean triple.

Hence set A will contain factors of 32 i.e. {1,3,9}

Solution set for D will be {2x (1,3,9) U 2x (1,3,9) U 2x (1,3,9)}

i.e. {2,6,18,4,12,36,8} [since in last iteration, 23*3 (2nd element of set) = 24 we need to consider only till 1st element]. Hence there will be 7 triples having 24 as one of the sides.

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