I need to solve for $(n_1, n_2, m)$ (where $m, n_1, n_2$ are all integers - may be +ve or -ve) satisfying $m^2=n_1^2+n_1n_2+n_2^2$.
I have found 2 solutions so far - $(3, 5, 7)$ and $(7, 8, 13)$. What is a general solution?
UPDATE: For my purpose, it is further needed that $\gcd(n_1, m) = \gcd(n_2, m) = 1 $.
This equation is symmetric so many solutions.
Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.
Write the formula can someone come in handy. the equation:
$Y^2+aXY+X^2=Z^2$
Has a solution:
$X=as^2-2ps$
$Y=p^2-s^2$
$Z=p^2-aps+s^2$
more:
$X=(4a+3a^2)s^2-2(2+a)ps-p^2$
$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$
$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$
more:
$X=(a+4)p^2-2ps$
$Y=3p^2-4ps+s^2$
$Z=(2a+5)p^2-(a+4)ps+s^2$
more:
$X=8s^2-4ps$
$Y=p^2-(4-2a)ps+a(a-4)s^2$
$Z=-p^2+4ps+(a^2-8)s^2$
For the particular case: $Y^2+XY+X^2=Z^2$ You can draw more formulas.
$X=3s^2+2ps$
$Y=p^2+2ps$
$Z=p^2+3ps+3s^2$
more:
$X=3s^2+2ps-p^2$
$Y=p^2+2ps-3s^2$
$Z=p^2+3s^2$
In the equation: $X^2+aXY+bY^2=Z^2$ there is always a solution and one of them is quite simple.
$X=s^2-bp^2$
$Y=ap^2+2ps$
$Z=bp^2+aps+s^2$
$p,s$ - integers asked us.
I'll first address the gentleman INDIVID why he prefers, in over 90% of his answers, to give results without proofs or links/reference to sources. One can hardly learn from such methods even if the results are concrete. Given; $m^2=n_1^2+n_1n_2+n_2^2$. We have, $n_1^2+n_1n_2+n_2^2-m^2=0$, a quadratic in $n_1$, hence, for the discriminant we have, $n_2^2-4(n_2^2-m^2)=k^2$ or $(2m)^2-3n_2^2=k^2$ for some integer $k$. Now, all rational pell equation of the above form has solutions given by $(r^2-d)^2=(r^2+d)^2-d(2r)^2$ for any $r$. Hence, by direct comparison, $d=3$, $n_2=2r$, $2m=r^2+3$ and $k=r^2-3$ where $r$ is any odd integer, a condition which ensures $m$ is an integer. Now, $n_1$ must be an integer, since $k$ and $n_2$ have the same parity.
For you second question; given the integer $d$, any equation of the form $x^2-dy^2=z^2$ has the general, integer, solutions $x=r^2+d$, $y=2r$ and $z=r^2-d$ for any integer $r$
– user145342 Jun 01 '14 at 10:17Method $1$
The remarkable equation $x^2 + pxy + y^2 = z^2$ where $p$ is a fixed integer. The well-known pythagorean equation is obtained for $p = 0$, and your question for $p=1$.
The solutions of this equation has been completely determined, all integral solutions to this equation are given by: $$x = k(pn^2 −2mn)$$ $$ y = k(m^2 −n^2)$$ $$ z = k(pmn−m^2 −n^2)$$ where k,m,n are integral parameters.
Method $2$ We can notice that your equation is equivalent to $$(2n_1+n_2)^2+3n_2^2=(2m)^2$$ The solutions of the equation $x^2+3y^2=z^2$ are also completely determined, see for example here, otherwise there is a generalization here.
