# New way to find Pythagorean Triples

## 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 **x ^{2}** then we will always have integer values for Pythagorean Triples since D will get canceled from numerator & denominator Moreover D should be such that X

^{2}/ 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 = |(X ^{2} – D^{2})/2D|**

**Hypotenuses = (X ^{2} + D^{2})/2D**

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

Let set ‘A’ contain factors of y^{2}

**Solution set for D = {2 ^{1} * all members of set A U 2^{2} * all member of set A U … 2^{n} * all member of set A}**

(Here ‘U’ stands for union)

[Note:- While solving last part (i.e. 2^{n} * all member of set A), one should stop as soon as 2^{n} * 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 = 2^{0} * 35 hence y = 35.

Hence set A will contain factors of 35^{2} i.e. {1,5,7,25,35,49,175,245,1225}

However Solution set for D will be {1,5,7,25} [as 2^{0}*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 = 2^{3} * 3 hence y = 3.

3) if** X=64** then it means X = 2^{6} * 1 hence y=1

Hence set A will contain factors of 1^{2} i.e. {1}

Solution set for D will be {2^{1 }x (1) U 2^{2 }x (1) U 2^{3 }x (1) U 2^{4}^{ }x (1) U 2^{5 }x (1) U 2^{6 }x (1)}

i.e. {2,4,8,16,32} [since in last iteration, 2^{6}*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 = 2^{0} * 11 hence y=11

Hence set A will contain factors of 11^{2} i.e. {1,11,121}

Solution set for D will be {1} [as 2^{0}*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 3^{2} i.e. {1,3,9}

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

i.e. {2,6,18,4,12,36,8} [since in last iteration, 2^{3}*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.