It all started when I received a comment on
one of my older posts, asking me to prove
what I had stated in the post. This got me
thinking, as I myself had forgotten the method
by which I had gotten the result.
In the last few posts, I have tried to reconstruct
my solution, and improve and generalize it.
The posts have been about trying to generate
Pythagorean Triples, given an initial value.
Until now, the only triples I could generate,
where ones in which two of the numbers are
separated by either 1 or 2.
I have been trying recently to generate an
algorithm by which I could, given a number,
generate all the Pythagorean Triples in which
the number occurs. And I believe I have
succeeded. Below is my work. As I am
currently in my school dormitory, I cannot
upload any properly formatted equations, nor
scanned paperwork. I will try to do so on the
following weekends.
a2 + b2 = c2 [b = c - k]
a2 = c2 - b2
a2 = (c + b)(c - b)
a2 = (c + c - k)(c - c + k) [Substituting b for (c-k)]
a2 = (2c - k)(k) [Now we isolate c]
a2/k + k = 2c
(a2/2k) + (k/2) = c
(a2/2k) - (k/2) =b
This is the result we hoped for. However,
the equation seems slightly faulty at first
sight, due to the fact that there is division by
2. What if an odd number turns up? However,
this doesn't cause a problem because regardless
of whether k is odd or even, c is a whole
number, except when a2 is not divisible by k.
Therefore, the algorithm runs like this:
Find all factors of a2, and use them as differing
values of k. Then, for each value of k, use the
equation to find values of b and c. There will
always be at least one valid whole number
solution, but there might be other solutions,
not including fractions or ones where b or c
are 0. Below I demonstrate my method with
a few examples.
a = 6, a2 = 36. [k = 1, 2, 3, 4, 6, 9, 12, 18, 36]
k = 1: b = 17.5, c = 18.5
k = 2: b = 8, c = 10 [62 + 82 = 102]
k = 3: b = 4.5, c = 7.5
k = 4: b = 2.5, c = 6.5
k = 6: b = 0, c = 6
k = 9: b = (-)2.5, c = 6.5
k = 12: b = (-)4.5, c = 7.5
k = 18: b = (-)8, c = 10
k = 36: b = (-)17.5, c = 19.5
Here, only one cases of k, out of 9 worked
out. However, every value mentioned above
works, even if it is not an integer.
In the future, I hope to build on my older posts,
and also add on to this series by displaying
my maths, properly formatted by LaTeX.
one of my older posts, asking me to prove
what I had stated in the post. This got me
thinking, as I myself had forgotten the method
by which I had gotten the result.
In the last few posts, I have tried to reconstruct
my solution, and improve and generalize it.
The posts have been about trying to generate
Pythagorean Triples, given an initial value.
Until now, the only triples I could generate,
where ones in which two of the numbers are
separated by either 1 or 2.
I have been trying recently to generate an
algorithm by which I could, given a number,
generate all the Pythagorean Triples in which
the number occurs. And I believe I have
succeeded. Below is my work. As I am
currently in my school dormitory, I cannot
upload any properly formatted equations, nor
scanned paperwork. I will try to do so on the
following weekends.
a2 + b2 = c2 [b = c - k]
a2 = c2 - b2
a2 = (c + b)(c - b)
a2 = (c + c - k)(c - c + k) [Substituting b for (c-k)]
a2 = (2c - k)(k) [Now we isolate c]
a2/k + k = 2c
(a2/2k) + (k/2) = c
(a2/2k) - (k/2) =b
This is the result we hoped for. However,
the equation seems slightly faulty at first
sight, due to the fact that there is division by
2. What if an odd number turns up? However,
this doesn't cause a problem because regardless
of whether k is odd or even, c is a whole
number, except when a2 is not divisible by k.
Therefore, the algorithm runs like this:
Find all factors of a2, and use them as differing
values of k. Then, for each value of k, use the
equation to find values of b and c. There will
always be at least one valid whole number
solution, but there might be other solutions,
not including fractions or ones where b or c
are 0. Below I demonstrate my method with
a few examples.
a = 6, a2 = 36. [k = 1, 2, 3, 4, 6, 9, 12, 18, 36]
k = 1: b = 17.5, c = 18.5
k = 2: b = 8, c = 10 [62 + 82 = 102]
k = 3: b = 4.5, c = 7.5
k = 4: b = 2.5, c = 6.5
k = 6: b = 0, c = 6
k = 9: b = (-)2.5, c = 6.5
k = 12: b = (-)4.5, c = 7.5
k = 18: b = (-)8, c = 10
k = 36: b = (-)17.5, c = 19.5
Here, only one cases of k, out of 9 worked
out. However, every value mentioned above
works, even if it is not an integer.
In the future, I hope to build on my older posts,
and also add on to this series by displaying
my maths, properly formatted by LaTeX.
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