Find ALL Phthegorean Triples containing integer 12Solutionwe

Find ALL Phthegorean Triples containing integer 12.

we need to find all pyhthagorean Triples containing integer 12 :

first find all primitive Pythagorean triples containing a divisor of 12,

so after that by multiplying by an appropriate factor, we get Pythagorean tripes containing integer 12.

Since we know that there is only one Pythagorean triple containing 0, we are going to avoid those triples that contain 0.

we know that divisors of 12 are given by 1, 2, 3, 4, 6, 12.

now find primitive Pythagorean triples containing 1.

we know that

(x = m² n² , y = 2mn, z = m² + n²⁾ is primitive when only one of m, n is even and also they are relatively prime.

Since we have taken care of the ones with 0, we only want m > n > 0.

so if x = 1, then 1 = m² n² = (m n)(m + n)

as m, n are positive,

m n = 1 and m + n = 1 adding both of them we have 2m = 2 hence m = 1 subtracting both of them n = 0,

so this results in a = 0 as in the above case.

y is not = 1 since y is even and if z = 1, then m² + n²= 1, but both m, n > 0 so their sum is greater than 1 which is also not possible.

Now, find primitive Pythagorean triples containing 2.

as we have y which is the only even element, we know that 2 = 2mn and so m = 1 and n = 1.

This gives x = m² n² = 0 0 = 0 which is a case we have already considered

now find primitive Pythagorean triples containing 3

first take x = 3

s0 3 = m² n² = (m n)(m + n).

as m and n btoh are positive we set m + n = 3 and m n = 1.

solve these equation

by adding both of them gives 2m = 4 hence m = 2

by subtracting both of them gives 2n = 2 hence n = 1

so we get (3, 4, 5) as a solution.

for z is not equal to 3 since 3 is not a sum of two squares.

now find primitive Pythagorean triples containing 4,

y is the only even element, we have 4 = 2mn and so 2 = mn.

as m > n, we can set m = 2 and n = 1.

which gives the same Pythagorean triple as above.

now find primitive Pythagorean triples containing 6,

as y is the only even element we have 6 = 2mn and so 3 = mn.

as m > n, we set m = 3 and n = 1.

as noth m and n are odd this does not correspond to a primitive triple

now find primitive Pythagorean triples containing 12,

y is the only even element we have 12 = 2mn or 6 = mn.

Since m > n, there are two possiblities:

m = 6 and n = 1

m = 3 and n = 2.

so the first gives triple as (x = 6² 1² = 35, y = 2 · 6 · 1 = 12, z = 6² + 1² = 37) = (35,12,37)

second gives a triple as (x = 3² 2² = 5, y = 2 · 3 · 2 = 12, z = 3² + 2² = 13). = (5,12,13)

Pythagorean triples are (0, 12, 12), 4 · (3, 4, 5) = (12, 16, 20), 3 · (3, 4, 5) = (9, 12, 15)., (5, 12, 13), and (35, 12, 37).

which is (0, 12, 12), (12, 16, 20), (9, 12, 15), (5, 12, 13), and (35, 12, 37).