Show that if a quadratic form is primitive then so are equivalent forms A Quadratic form is primitive if the greatest common divisor of the coefficients of it's terms is 1. 
I saw in number theory book that "it is easily seen that any form equivalent to a primitive form is also primitive"  but I cannot seem to show this to be true myself. 
(https://books.google.co.uk/books?id=njgVUjjO-EAC&pg=PA140&lpg=PA140&dq=showing+equivalent+forms+are+primitive&source=bl&ots=ckh4-Wior1&sig=3xcVHyXRjPGci63-RWnmjmFM2aw&hl=en&sa=X&ved=0ahUKEwix_5OWlZbMAhXIaxQKHdA_A4YQ6AEINDAD#v=onepage&q&f=false -between the first and second definition on the linked page)
Any help with what I assume is a very simple proof would be much appreciated!
 A: If $q(x,y)=ax^2+bxy+cy^2$ then $$\gcd(a,b,c)=\gcd(a,a+b+c,c)=\gcd(q(1,0),q(1,1),q(0,1)).$$
If moreover $q(x,y)$ equals $Q(X,Y):=AX^2+BXY+CY^2$ with $X=\alpha x+\beta y$ and $Y=\gamma x+\delta y$ then
$\gcd(a,b,c)=\gcd(\text{three values of }Q)$ is a multiple of $\gcd(A,B,C)$.
And conversely, if your change of variables is invertible.
A: Beginning with $\langle A,B,C \rangle,$ all equivalent forms can be created by a sequence of operations of simple types. Convenient to use three:
$$\langle A,B,C \rangle \mapsto \langle C, \; -B, \; A \rangle,  $$
$$\langle A,B,C \rangle \mapsto \langle A, \; B+2A, \; A+B+C \rangle,  $$
$$\langle A,B,C \rangle \mapsto \langle A, \; B-2A, \; A-B+C \rangle.  $$
In all three cases, you can check that $\gcd(A,B,C) = 1$ means that the three new coefficients are coprime as a triple.
There are more elegant ways to present the modular group, but this is closest to what you would do by hand. 
Note that equivalence means this: take the Hessian matrix $H$ of second partial derivatives of your form, so that, for a column vector $x,$ $f(x) = (1/2) x^T H x.$ Then, given a matrix $P$ of integers and determinant $1,$ the Hessian matrix of the transformed item is $G =P^T H P.$
