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Consider the equation $27x + 14y + 10z = 1$. Give parameterized solutions for all integer solutions $x$, $y$, $z$.

How do I find integer solutions?

Any help appreciated.

Well, $gcd(27, 14) = 1$

So we can solve $27x + 14y = 1$. And on such solution is $x=-1$ and $y = 2$. We can generalize all possible solutions as $x = -1 + 14m$ and $y = 2 - 27m$.

So for any $k$ we can find a solution to $27x + 14y = 1$ as $x = -k$ and $y = 2k$ and we can generallize all possible solutions as $x = -k + 14m$ and $y = 2k - 27m$.

And for any $z$ we can find solutions to $27x + 14y = 1- 10z$ or $27x + 14y + 10z = 1$ by $x = 10z -1 + 14m$ and $z = 2(1-10z) - 27m$.

So the set of solutions are that $(x,y,z) = (10v -1 + 14m, -20v + 2 - 27m, v)$ for any integers $v, m$.

BLURB: one may find tons of stuff on LLL reduction of a lattice basis. It is pretty much impossible to find material on how to find a basis in the first place, where we are given an integral lattice of small rank in Euclidean space and asked to find a basis for the orthogonal lattice.

original:

It is not so clear how to find a basis for the orthogonal lattice of a given integer lattice. However, there is a method that you either know already or will learn in linear algebra. We just need to stick with integer elements of all matrices. We want $$ (27,14,10) P = (1,0,0) $$
where $P$ is a square matrix with integer elements and determinant $1.$ It took me a few steps, that is I did four column operations and took $P = P_1 P_2 P_3 P_4$ where each $P_j$ is one of those "elementary " matrices.

The out come is
$$
left(
begin{array}{ccc}
27&14&10
end{array}
right)
left(
begin{array}{ccc}
1&-14&-10 \
1&-13&-10 \
-4&56&41
end{array}
right)=
left(
begin{array}{ccc}
1&0&0
end{array}
right)
$$
It follows that, as columns, a basis for the orthogonal lattice is given by
$$
B^T =
left(
begin{array}{rr}
-14&-10 \
-13&-10 \
56&41
end{array}
right)
$$
This is not a reduced basis, the Gram matrix is
$$
BB^T =
left(
begin{array}{rr}
3501&2566 \
2566&1881 \
end{array}
right)
$$
However, the determinant is fairly small, $1025.$

It is not difficult to create a reduced basis in small dimension
$$
left(
begin{array}{rr}
-14&-10 \
-13&-10 \
56&41
end{array}
right)
left(
begin{array}{rr}
-3&8 \
4&-11 \
end{array}
right) =
left(
begin{array}{rr}
2&-2 \
-1&6 \
-4&-3
end{array}
right)
$$
Note that $ 33-32=1$

The new Gram matrix is
$$
BB^T =
left(
begin{array}{rr}
21&2 \
2&49 \
end{array}
right)
$$
Beginning with the solution $(1,1,-4)$ for $27x+14y+10z = 1,$ we get all solutions as
$$ (1+2m-2n, 1-m+6n, -4-4m-3n) $$

$27 cdot 3 + 10cdot (-8) = 1\
27cdot (-1) + 14cdot 2 = 1\
27cdot 2 + 14cdot(-1) + 10cdot(-4) = 0$

$x$ must be odd

let $x = 2n-1$

Then $y = -n+2, z = -4n$ will be solutions.

And we can add in another degree of freedom trading y's for z's

$(x,y,z) = (2n-1, -n+2+5m, -4n-7m)$