Exercise 3.5.14* (There is an integer $n$ properly represented by $f$ with the property that $(n,k) = 1$)

Proof. Consider the primitive form $f(x,y)=a{x}^2+\mathit{bxy}+c{y}^2$, where $a\wedge b\wedge c=1$. Let $k$ be a nonzero integer. Since $n\wedge k=n\wedge (-k)$, we can suppose that $k\ge 1$.

We show first that if $k=p$ is a prime number, there is some $n$ represented by $f$ which is not a multiple of $p$. If not, then for all $(u,v)\in {\mathbb{Z}}^2$,

Then $a=f(1,0)\equiv 0(\mathit{mod}p)$, $c=f(0,1)\equiv 0(\mathit{mod}p)$, and $a+b+c=f(1,1)\equiv 0(\mathit{mod}p)$, thus $b=(a+b+c)-a-c\equiv 0(\mathit{mod}p)$. This contradicts the fact that $f$ is a primitive form. Therefore $f$ represents $p$.

If $k=1$, any integer $n$ primitively represented by $f$ (for instance $n=f(1,0)$) is suitable. We suppose now that $k>1$. Let $k=p_1^{a_1}\cdots p_l^{a_l}$ be the decomposition of $k>1$ in prime factors.

For each prime divisor $p_i$ of $k$, there are integers $u_i,v_i$ such that

By the Chinese Remainder Theorem, there is an integer $u$ such that $u\equiv u_i(\mathit{mod}{p}_i)(1\le i\le l)$, and an integer $v$ such that $v\equiv v_i(\mathit{mod}{p}_i)(1\le i\le l)$. Then

This shows that $m=f(u,v)$ is represented by $f$, and $m\wedge k=1$. But this representation is not necessarily primitive.

Note that $(u,v)\ne (0,0)$, otherwise $m=f(u,v)=0$, which is impossible since $m\not\equiv 0(\mathit{mod}{p}_1)$. Therefore $d=u\wedge v\ne 0$. If $x_0=u∕d,y_0=v∕d$, then $x_0\wedge y_0=1$. Moreover $m=d^{2}f(x_0,y_0)$, thus $n=m∕d^2$ is an integer, $n\wedge k=1$, and $n=f(x_0,y_0)$ is properly represented by $f$.

In conclusion, if $k$ is a nonzero integer, then there exists an integer $n$ properly represented by $f$ with the property that $n\wedge k=1$. □