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!
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 $G =P^T H P.$