Find the complete and singular integrals of p^2+q^2 = 4z

Find the complete and singular integrals of \[ p^2+q^2 = 4z. \]

Explanation of Method

Since $p^2+q^2=4z$ suggests $z$ is a perfect square of a linear form, try $z=(ax+by+c)^2$ with the normalisation $a^2+b^2=1$ needed for consistency; the singular integral follows from $F_p=F_q=0$.

Step 1 · Try $z=(ax+by+c)^2$

Then $p=2a(ax+by+c)$, $q=2b(ax+by+c)$, so \[ p^2+q^2 = 4(a^2+b^2)(ax+by+c)^2 = 4z \] requires $a^2+b^2=1$.

Step 2 · Find the singular integral

With $F=p^2+q^2-4z$: $F_p=2p=0$, $F_q=2q=0$ give $p=q=0$; substituting into $F=0$ gives \[ z = 0. \]

Verification

Complete integral: $p^2+q^2=4(a^2+b^2)(ax+by+c)^2=4z$ when $a^2+b^2=1$. \checkmark
Singular integral: $p=q=0$ gives $p^2+q^2=0=4\cdot0$. \checkmark

Result

The complete and singular integrals are:

\[ \boxed{ \text{Complete: } z=(ax+by+c)^2,\ a^2+b^2=1; \qquad \text{Singular: } z=0 } \]