If $ac-bd=p$ and $ad+bc=0$, then $a^2+b^2\neq 1$ and $c^2+d^2\neq 1$? I'm trying to prove the following:

Let $a,b,c,d\in\Bbb{Z}$ and $p$ be a prime integer. If $ac-bd=p$ and $ad+bc=0$, prove that $a^2+b^2\neq 1$ and $c^2+d^2\neq 1$.

Actually I'm not even sure if this is correct. A proof or counter-example (in case this assertion is wrong) would be great. 
I got the following 3 results:


*

*$p^2=(a^2+b^2)(c^2+d^2)$

*$b(c^2+d^2)=-pd$

*$a(c^2+d^2)=pc$

*$c(a^2+b^2)=pa$

*$d(a^2+b^2)=-pb$


I'm at a loss as to how to proceed beyond this. Assuming $c^2+d^2=1$ or $a^2+b^2=1$ does not seem to cause any contradictions. 
Thanks in advance!
 A: Consider $p^2 = p^2 + 0^2 = (ac-bd)^2 + (ad+bc)^2 = (a^2+b^2)(c^2+d^2)$. Then either one of $a^2+b^2$, $c^2+d^2$ equals $p^2$ and the other equals $1$ or both equal $p$. This follows from unique factorization of $\mathbb{Z}$.
Considering the first case, suppose wlog $a^2+b^2=1$ and $a=0$. Then $c^2+d^2=p^2$. Now if both $c$ and $d$ are nonzero, then, since $b$ is also not equal to zero, $ad+bc$ cannot be zero, a contradiction. Thus suppose, again wlog, that $c=0$ and $d^2=p^2$. This leads to a counterexample, namely $a=0,b=1,c=0,d=-p$ or $a=0,b=-1,c=0,d=p$.
A: The idea is that there aren't a lot of ways we can have $a^2 + b^2 = 1$ or $c^2 + d^2 = 1$ in the first place! 
If $a^2 + b^2 = 1$, then one of $a, b$ is zero and the other has absolute value $1$. Let's say $a = 0$ and $b = \pm 1$. Then $ac - b d = -b d = p$, so $d$ is $\pm p$. But we also have $ad + b c = b c = 0$, so $c = 0$. 
And now we've found a counterexample that shows the problem statement is false: take $a = 0, b = 1, c = 0, d = -p$. Notice we have $a^2 + b^2 = 1$.  
