At 3:53 PM -0500 11/30/97, Paul E. Miller wrote:
>mark snyder wrote:
>
(snip)
>> The PPTs [since at least one of a and b has to be even, choose it to be b]
>> can be parametrized by the following:
>>
>> a = m^2 - n^2,
>>
>> b = 2mn,
>>
>> c = m^2 + n^2
>>
>> where m and n are relatively prime integers of opposite parity (if m is
>> even, n is odd, and vice-versa).
(snip)
>> mark snyder
>>
>Could this parameterization also be used to prove the following:
>If a, b, and c are a PT, then 3 is a factor of a, b, or c, AND 4 is a
>factor of a, b or c, AND 5 is a factor of a, b, or c?
>
>Paul Miller
>Parkersburg WV
Yes, and even more than that. In fact, this is true of PPTs. If a, b, c
are a PPT, then
3 is a factor of either *a or b*,
5 is a factor of a, b, or c, and
since one of m and n must be even, b must be divisible by 4
As a consequence, the product abc must be divisible by 60.
Of course, if it holds for PPTs, the same result must hold for PTs.
Each of these results is pretty accessible to proof by HS students who feel
comfortable with factoring polynomials.
For instance one way of showing that either a or b is divisible by 3 goes
like this:
The product ab is
2mn(m^2-n^2).
This can be rewritten as (ignoring the 2; we know *that* is not divisible
by 3!):
mn(m^2-n^2) = mn[ (m^2-1) - (n^2-1) ] = n{ m(m^2-1) } - m{ n(n^2-1) }
=n{ (m-1)m(m+1) } - m{ (n-1)n(n+1) }.
The terms in {} are just a product of three consecutive integers, hence are
each divisible by 3!=6. Hence, the product of a and b is divisible by 6,
and hence by 3. Since 3 is a prime, if it divides ab, it must divide one
of a or b. QED
A somewhat more involved proof goes through for 5, using the product abc.
I'm not interested in starting any flame wars here, but how would you go
about showing this with a TI-8x?
mark snyder
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