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# Pythagorean Triplets

**ab**

We seek to identify all positive integers *a*, *b*, and *c* so that
*a*^{2}+*b*^{2}=*c*^{2}. A triple (*a*,*b*,*c*) satisfying this equation is
called a *pythagorean triplet.* Since any multiple
of a pythagorean triplet is again one, we only seek *
primitive* ones where *a*, *b*, and *c* have no common factor.
By setting
and ,
we can first seek
rational numbers *x* and *y* so that *x*^{2}+*y*^{2}=1.
=2in
Now, if (*x*,*y*) is such a pair of rational
numbers, the line from (*x*,*y*) to (-1,0) will have rational
slope - call it *t*. Along this line, *y*=*t*(*x*+1), so
*x*^{2} +
*t*^{2}(*x*+1)^{2}=1 telling us that

(1+*t*^{2})*x*^{2} + 2*t*^{2} *x* = 1-*t*^{2}

or

Complete the square on the above equation giving

Taking square roots gives

Write
with *u*, *v* positive integers having no common
factor. This gives

Using
and
we see that

for some rational number *k*.
By thinking about the highest power *p*^{r} of
an odd prime appearing in the denominator of *k* (written in lowest terms),
one sees that *p*^{r} divides *uv*, *u*^{2}+*v*^{2}, -*u*^{2}+*v*^{2}, 2*u*^{2}, and 2*v*^{2}.
This means
divides both *u* and *v* contradicting *u*
and *v* having no common factor, UNLESS *r*=0. Similarly by
noting exactly one of *a* and *b* can be odd, and choosing *b* to
be the even number, we can eliminate powers of 2 in
the denominator of *k*. This shows that *k* is
a whole number and we obtain:
**Theorem:** The pythagorean triplets with *b* even
are exactly the triples of the form

where *u* and *v* are positive whole numbers with no common factor.

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*root*

*2002-09-26*