# Does the function which sends a right angled triangle to its area produce infinitely many numbers having hardly any prime factors?

Let $T$ be the set of pythagorean triples, that is, triples of integers (a,b,c) satisfying a2 + b2 = c2. We think of $T$ as the set of right angles triangles with integer lengths. And let $f : T \rightarrow \mathbb{Z}$ be the function $(a,b,c) \mapsto \frac{ab}{12}$ which computes the area of a triangle (divided by 6, which seems to always be a factor for some reason).

I was wondering: what are the number theoretic propertires of $f$? It seems to produce numbers with few prime factors. What is the reason for this? For instance, $f(3,4,5) = 1$, $f(36,77,85) = 3 * 11 * 7$, and $f(65,72,97)=39*5*2$. Can we put a bound on the number of prime factors in the numbers that $f$ spits out? Or at least, can we give a 'generic' statement such as 'The output of $f$ almost always spits out numbers with less than 8 factors' or something?

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ab is divisible by 12 because the only Pythagorean triple mod 3 or mod 4 is 1^2 + 0^1 = 1^2. In any case, since we know that a = 2kmn, b = k(m^2 - n^2) for some integers k, m, n, f can produce numbers with arbitrarily many prime factors. – Qiaochu Yuan Jul 14 '10 at 19:57
Also, bear in mind that in general most numbers have very few prime factors; it would be notable if the function consistently produced 'round' numbers with many small prime factors. – Thomas Bloom Jul 14 '10 at 22:32
(Also, 39 is not prime...) – Qiaochu Yuan Jul 14 '10 at 23:24
Prop: 39 is divisible by 57. Proof: It is not relatively prime to 57, and by [Grothendieck], 57 is a prime. $\square$ – Victor Protsak Jul 15 '10 at 0:42
Standard sieve methods will produce a (finite!) integer $d$ such that there are $\gg_{\varepsilon} T^{2-\varepsilon}$ Pyathagorean triples $(a,b,c)$ with $0<a,b<T$ and $ab$ having at most $d$ prime factors. – David Hansen Jul 15 '10 at 2:58

Presumably, there are infinitely many primes $p$, $q$, such that $p+q$ is 6 times a prime and $p-q$ is 4 times a prime (e.g., $73+5=6\times13$, $73-5=4\times17$). This ought to follow from the prime $k$-tuples conjecture. Proving it is another matter.

Slightly simpler (but still out of reach), there should be infinitely many primes $p$, $q$, such that $2p+q$ is prime and $2p-q$ is 3 times a prime (e.g., $p=13$, $q=5$). Then the integer right-triangle with sides $4pq$ and $4p^2-q^2$ has area 6 times a product of 4 primes.

EDIT: Maybe not prime $k$-tuples but Schinzel's Hypothesis H is what's needed.

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Thanks Gerry, very handy! – Bruce Bartlett Jul 15 '10 at 5:38
Ah, I see there is a proof of something like this in 'Equidistribution and Primes' by Peter Sarnak... – Bruce Bartlett Jul 16 '10 at 11:42

There are many right triangles whose area has as few prime factors as possible:

The Green-Tao paper "Linear equations in primes", and the subsequent work on their Mobius Nilsequences conjecture by GT and Gowers Inverse conjecture by GT-Ziegler, implies that $\frac{ab}{12}$ is infintely often a product of four primes, in a quantitative sense. Indeed, reparametrizing gives $ab=12xy(2x+3y)(2x-3y)$, and now we are asking for points in the lattice $(1,0,2,2)\mathbb{Z}+(0,1,3,-3)\mathbb{Z}$ all of whose coordinates are prime. This system has finite complexity and thus the main results of "Linear equations in primes" applies unconditionally, that is to say the number of pairs $(x,y)$ with $0 < |x| , |y| < T$ and $xy(2x+3y)(2x-3y)$ a product of four primes is asymptotically $cT^2 (\log{T})^{-4}$ for some constant $c$.

This deduction can be found, for example, in Sarnak's notes "Equidistribution and primes".

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