We know that there is a famous theorem to determine whether there is an integer solution to the Legendre-Diophantine equation with three variables.
Legendre theorem of ternary quadratic form
The indeterminate (Diophantine) equation
$$ax^2+by^2+cz^2=0$$
whose coefficients a, b and c are pairwise coprime integers, square-free and not all of the same sign, has a non-zero rational solution if and only if all following three congruences are solvable: \begin{align*} u^2\,&\equiv-bc\pmod{|a|},\\ v^2\,&\equiv-ca\pmod{|b|},\\ w^2&\equiv-ab\pmod{|c|},\\ \end{align*}
https://encyclopediaofmath.org/wiki/Legendre_theorem
However, I was baffled by a Legendre equation $x^2 + 5 y^2 = 4389 z^2$.
(ps: $4389=3\cdot7\cdot11\cdot19$)
With the help of the above theorem and software, we found by error that the Diophantine equation $\text{exists}$ a non-zero integer solution, but I applied the software again to search brutally, and found no solution less than $10000$.
[Edit]In the process of judging whether the congruence equation has a solution, I mistakenly think that Jacobian sign is equivalent to Legendre sign.
Now I want to know how to estimate the magnitude of least positive integer solution for this Diophantine equation?
Wolfram Mathematica code:
In[1]={JacobiSymbol[-a*b, Abs[c]],
JacobiSymbol[-a*c, Abs[b]],
JacobiSymbol[-b*c, Abs[a]]} /. {
a -> 1, b -> 5, c -> -4389}
Out[1]={1, 1, 1}
In[2]=Solve[x^2 + 5 y^2 == 4389 z^2 && 0 < {x, y, z} < 10000, {
x, y, z}, Integers]
Out[2]={}
Edit
$$x^2 + 5 y^2 = 20769 z^2$$
Thanks for Jagy's solution. I also want to know the magnitude between the least positive integer solution and the coefficient of the above Diophantine equation.
Solve[x^2 + 5 y^2 == 20769 z^2 && 0 < {x, y, z} < 100, {
x, y, z}, Integers]
{{x -> 17, y -> 64, z -> 1}, {x -> 58, y -> 59, z -> 1},
{x -> 82, y -> 53, z -> 1}}
well, no, Legendre symbol $$ (-20|11) = (-1|11)(5|11) = (-1|11)(11|5) = (-1|11) = -1 $$
If $x^2 + 5 y^2 $ is divisible by $11,$ it must be divisible by $121 = 11^2.$ Indeed, writing $$11^t \parallel a^2 + 5 b^2 $$ to mean $11^t | a^2 + 5 b^2 $ but not $11^{t+1} |a^2 + 5 b^2, $ the exponent $t$ must be even. Writing $t=2 \tau,$ we must have $11^\tau | a $ and $11^\tau | b $
You need to check each prime that divides one of the coefficients
from page 82, Corollary 1, in Cassels, Rational Quadratic Forms: if $abc$ is squarefree, while $ax^2 + b y^2 + c z^2$ is indefinite, such that there is a nonzero integer solution to $ax^2 + b y^2 + c z^2=0,$ then there is such a solution with $$|a|x^2 + |b| y^2 + |c| z^2 < 4 |abc|.$$ Note that allowing square factors in $abc$ would raise the size of the smallest solution.
IF $$ x=101 u^2 -492uv -429v^2 \; ,$$ $$ y=188 u^2 +318uv +24v^2 \; ,$$ $$ z=3 u^2 +4uv +3v^2 \; , $$ then $$ x^2 + 5y^2 = 20769 z^2 \; . $$ which gives infinitely many solutions, including some small ones.
More matrices of coefficients of $u^2,uv,v^2$ The first one gives $$ x=17 u^2 -640uv -429v^2 \; ,$$ $$ y=64 u^2 +34uv +24v^2 \; ,$$ $$ z= u^2 + 5v^2 \; . $$
Umm. There are several solutions with $z=1,$ since
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