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Add $p=167$ example, plus a few small corrections and explanation while I'm at it
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Noam D. Elkies
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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 equationequations, as Gerhard "insert quote here" Paseman suggested; but the difficulty is not the size of the fundamental unit (which is can be as small as $p + \sqrt{p^2-1}$) but but the rare and unpredictable appearance of primes in the the resulting sequence; I doubt that anything can be proved about the question. [The Diophantine equation for a triangle of sides $p,q,q-1$$p,q,q-d$ to be Heronian is $(p^2-1) (2q-p-1) (2q+p-1) = x^2$$(p^2-d^2) (2q-p-d) (2q+p-d) = x^2$.]

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). This comes to a Fermat-Pell equation, as Gerhard "insert quote here" Paseman suggested; but the difficulty is not the size of the fundamental unit (which is $p + \sqrt{p^2-1}$) 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 $p,q,q-1$ to be Heronian is $(p^2-1) (2q-p-1) (2q+p-1) = x^2$.]

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 $(p^2-d^2) (2q-p-d) (2q+p-d) = x^2$.]

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Noam D. Elkies
  • 79.9k
  • 15
  • 281
  • 376

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). This comes to a Fermat-Pell equation, as Gerhard "insert quote here" Paseman suggested; but the difficulty is not the size of the fundamental unit (which is $p + \sqrt{p^2-1}$) 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 $p,q,q-1$ to be Heronian is $(p^2-1) (2q-p-1) (2q+p-1) = x^2$.]