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It is known that if $x^2 + y^2 = z^2$ is a primitive Pythagorean triplet then $z$ is not divisible by any prime of the form $4k-1$. The following is a generalization of this classical result which shows that the source of this property is not the hypotenuse $z$, but the two orthogonal side $x$ and $y$:

Conjecture: Let $f(x,y) = a_0x^n + a_1 x^{n-1}y + a_2x^{n-2}y^2 + \cdots + a_ny^n$, $n \ge 2, a_0a_n \ne 0$. Then there are infinitely many primes of the form $8k+3$ which do not divide $f(x,y)$ for any primitive Pythagorean triplet $x^2 + y^2 = z^2$.

Questions:

  1. Is the conjecture known? I am looking for a proof or disproof of the conjecture.
  2. Or can we prove the simpler case for $x^n +y^n$?

Update 1: If has been proved for the special case $x+y$ (in the answer below) and is already known to be true for $x^2 + y^2$. Experimental data shows that $x^3 + y^3$ is not divisible by infinitely many primes of the form $8k+3$ while $x^4 + y^4$ is not divisible by infinitely many primes of the form $8k+3, 8k+5$ and $8k+7$.

Update 2: Posted in MO since the general case of the conjecture is open

Sagemath Code

r = 2
fac = prime_factors(1)

while r <= 200:
    s = 1 + r%2
    while(s < r):
        if gcd(s,r)== 1:
            b = r^2 - s^2
            c = 2rs
            # t = 5b^3 +7b^1c^2 + 5b^2c^1 + 2c^3
            # t = b^4 - 3b^1c^3 - b^2c^2 - 1c^4
            # t = b^2 - 11bc - c^2
            t = b + c
            fac = fac + prime_factors(t)
            fac = list(dict.fromkeys(fac))
        s = s + 2
    r = r + 1


fac  = sorted(fac)
fac2 = fac
fac  = fac[:floor(0.5*len(fac))]


P = Primes()
prime = P[:prime_pi(max(fac))]


diff = list(set(prime) - set(fac))
diff = sorted(diff)
print diff
Nilotpal Sinha
  • 24,086

1 Answers1

7

For a primitive Pythagorean triangle, there are integers $m,n > 0$ with $\gcd(m,n) = 1$ and $m,n$ not both odd, so that $x+y$ equals $$ m^2 + 2mn - n^2 $$ This is an indefinite binary quadratic form of discriminant $8,$ which has class number one. Thus the form is equivalent to $$ u^2 - 2 v^2, $$ and any prime $$ q \equiv 3,5 \pmod 8 $$ is not able to divide such a form without dividing both variables. The special case of prime $2$ is taken care of by the condition that $m+n$ be odd.

Will Jagy
  • 146,052
  • This for the cases $f(x,y) = x+y$ ? For $f(x,y) = x^2 + y^2$ it is known to be divisible by primes of the form $8k+5$ – Nilotpal Sinha Jun 26 '20 at 02:48
  • @NilotpalKantiSinha yes, this is $x+y.$ Note that things would change just by throwing in constant coefficients, $x + 3y$ would not have the same problem primes as $x+y$ – Will Jagy Jun 26 '20 at 02:53
  • Your answer plus the fact that $x^2+y^2$ is not divisible by infinitely many primes of the form $4k−1$ can prove that there are infinitely many primes which do not divide $x^n+y^n$. Take the infinitely many primes which do not divide both $x+y$ and $x^2+y^2$. They will not divide $(x+y)(x^2+y^2)=x^3+y^3+xy(x+y)$. But we have no restriction on $x$ and $y$ so for every prime $p$ we will find some $x$ and $y$ such that $p|xy$. Hence the only option is that there are infinitely many primes which do not divide $x^3+y^3$. Similarly we can prove for higher powers. – Nilotpal Sinha Jun 26 '20 at 12:08