| Pythagorean Triples |
A little more fun with Pythagorean triples:
Exactly four right triangles with integer side lengths exist with a leg equal to 15. These are:
(15, 112, 113), (15, 20, 25), (15, 36, 39), and (8, 15, 17).
How many different right triangle with integer side lengths exist with a leg equal to 42? How do you know?
BONUS QUESTION:
42 is the product of 3 different prime numbers - 2, 3, and 7.
Resolve this problem given a leg which is the product of n different primes, given that, as in this problem, one of these primes is 2. |
Problem Moderated by: Sasha |
| Problem Solution |
The problem: how many pairs of integers (j , k) with k > j exist such that 422 + j2 = k2?
422 = k2 - j2 = (k + j)(k - j)
Since k and j are integers and k > j, k + j and k - j must be positive integers which, when multiplied, equal 422. This seems straightforward - find the number of factors of 422, subtract 1 (since we can't have k = 0), and divide by two.
42 = (2)(3)(7)
422 = 223272
So 42 has (2+1)3 = 27 factors, and 13 factor pairs.
However, not all factor pairs will give us integer values of j and k. Allow the two factors to be F1 and F2 so that F2 > F1 and F1F2 = 422
k + j = F2
k - j = F1
k = F1 + F2/2
j = F2 - F1/2
We can therefore allow both factors to be odd or both factors to be even, but we can never have one odd and one even. This means we must eliminate all cases where F1 OR F2 = 4k, where k is any factor of 3272. There are 9 such factors, so only 4 right triangles with integer side lengths and a leg equal to 42 exist. In general, we eliminate 3n-1 factor pairs. This leads us to our general equation for the bonus question -- 1/2(3n - 1) - 1/33n = 1/6(3n-3)
Incidentally, this proves that there are no right triangles with integer side lengths and a leg equal to 2 (that is, for n = 1, 1/6(3n - 3) = 0 ).
For a very interesting approach to this problem, please check out Graeme's solution at
http://mcraefamily.com/MathHelp/GeometryPythagoreanTriplesWithProductOfPrimesLeg.htm |
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