Heronian triangle with two sides that are prime Can any prime number form a Heronian triangle with a second prime as another side? I cannot find a second prime to form a Heronian triangle with either 23 or 167. I have checked up to the 10^7th prime for both with no solution. Also if there is a solution for primes other than 23 or 167 to have a second prime as another side, is the solution set for that prime finite or infinite. See OEIS A230666 and A233232 for primes 3 and 5 where the solution sets are infinite.
 A: For $p=23$, one Heronian triangle with another prime side $q$ has
$q = 5280071830550089$, with third side $q-1$ and area
$60663406817631420 = 2^2 \, 3^4 \, 11 \; 23 \; 37^2 \, 47 \; 71 \; 179 \; 181$.
For each $p$ there are probably infinitely many examples
but very sparse; for $p=167$ I didn't find one with a prime side
among the first few dozen solutions (though there might be a few other
variants to try).  [Added later: an example is
$q = 231781748893580717709514473745694370721$,
for which the triangle with sides $167$, $q-25$, $q$
has area $19135685576510124949571252858502010748400$
$$
= 2^4 \, 3 \; 5^2 \, 29 \; 43 \; 71 \; 167 \; 769 \; 29063 \; 250233481 \; 1154762937707.]
$$
This comes down to a few Fermat-Pell equations, as
Gerhard "insert quote here" Paseman suggested; but the difficulty
is not the size of the fundamental unit (which can be as small as 
$p + \sqrt{p^2-1}$) but the rare and unpredictable appearance of primes in 
the resulting sequence;
I doubt that anything can be proved about the question.
[The Diophantine equation for a triangle of sides $p,q,q-d$ to be Heronian is
$(p^2-d^2) (2q-p-d) (2q+p-d) = x^2$.]
