| Editorial corrections:
8. Sides of 160,163, and 3 do not make a triangle in my kind of universe.
I meant to say 160,162, and 3. Even that misses being an integral triangle by
a few thousandths. Anyway, in the meantime, I found an integral triangle
with sides 3,865, and 866. Very acute. An area of 1224.
9. This question should have the semi-universal bulletin board smile figure
appended to it, like so: :-)
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| 1) Given: Sides a, b, and c of a triangle, and its area, A, are all
integers.
Prove: A is even.
One formula for the area of a triangle is:
A = sqrt( s * (s-a) * (s-b) * (s-c) ),
where s = (a+b+c)/2. Call the product inside the square root function
P. Since P = A*A, P is an integer.
Lemma: s is an integer.
Suppose s be not an integer (this is proper English, see note
68 in SUMMIT""::SYS$NOTES:JOYOFLEX). Then s is in the form
i0/2, where i0 is an odd integer (i0 = a+b+c). Clearly, s-a,
s-b, and s-c also have forms i1/2, i2/2, and i3/2, for some
odd integers i1, i2, and i3. So P = i0*i1*i2*i3/16. But
i0*i1*i2*i3 must be odd, so P is not an integer, which is a
contradiction. Therefore s is an integer.
Lemma: P is even.
Suppose P be odd. Then none of s, s-a, s-b, and s-c could
be even integers. Since they are all integers, they must all
be odd. If s is odd and s-a is odd, then a must be even.
Similarly, b and c must be even.
If they are all even, then a = 2i, b = 2j, and c = 2k, for some
integers i, j, and k. Expanding P reveals
P = 2(i*i*j*j + j*j*k*k + k*k*i*i) - i**4 - j**4 - k**4.
Consider the remainder when P is divided by 4 when i, j, and k
take on even or odd values. Because i, j, and k are symmetric
in the above formula, it is only necessary to consider one case
where they are all even, one case where one is odd, one case
where two are odd, and so on.
i j k | formula, shown as congruences mod 4 | P mod 4
0 0 0 | 2(0+0+0)-0-0-0 | 0
0 0 1 | 2(0+0+0)-0-0-1 | 3
0 1 1 | 2(0+1+0)-0-1-1 | 0
1 1 1 | 2(1+1+1)-1-1-1 | 3
So P is congruent to either 0 or 3 modulo 4. If it is
congruent to 0, it is even, which contradicts the supposition
that P is odd. If it is congruent to 3, then A is not an
integer, because there is no integer whose square is congruent
to 3 modulo 4 (integers congruent to 0, 1, 2, and 3 have
squares congruent to 0, 1, 0, and 1, modulo 4). This is also
a contradiction.
Therefore the supposition that P is odd is false.
Since P is an even integer and A is an integer, then A is also even
(otherwise squaring A, an odd integer, would yield P, an even integer).
2) Is a Pythagorean triangle a right triangle with integral sides? If
so, one of the legs is necessarily even, so the area is obviously an
integer.
To see that one side is even, suppose they be both odd. Let the length
of the hypoteneuse be denoted by a and the lengths of the legs by b
and c. Then a*a = b*b + c*c. Since b and c are odd, b*b and c*c are
odd, and b*b + c*c is divisible by two but not by four. Clearly,
no integer has a square divisible by two but not by four. But a is
known to be an integer, so this is a contradiction. Therefore b and
c are not both odd.
-- edp (WHOAREYOU note 329)
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| Of course you all know the formula for generating the sides of
all Pythagorean triangles:
k(2mn)
k(m^2-n^2)
k(m^2+n^2)
where gcd(m,n)=1, m>n.
For quite a while, I have been researching the question of whether
there is a formula that generates all Integral Triangles
(also known as Heronian Triangles in the literature). Dickson
hedges the question in his monumental work, the History of the Theory
of Numbers. I have found several purported formulas in the literature,
some of which are wrong. I believe the following is a valid formula:
kmn(p^2+q^2)
kpq(m^2+n^2)
k(mq+np)(mp-nq)
where gcd(m,n)=1 and gcd(p,q)=1, mp>nq,
although the published proof seemed wrong.
Note that letting p=q=1 yields the formula for Pythagorean triples.
I hope you find this formula useful. Let me know if it misses any solutions.
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