Question: If the difference of the roots of $x^2-px+q=0$ is unity then prove that $p^2-4q=1$ and $p^2 =4 q^2=({1+2q})^2$

What I Tried

1.I proved the first part of the question using the understanding of the difference of the roots
$$D^{1/2}/|a|=|a-b|$$
for $ax^2+bx+c=0$ and D=Discrimininant

2. Do not know where to start.

Answer: 
You cannot prove it because second part is wrong. For example, the roots of $x^{2}-x=0$ are $0$ and $1$ so the hypothesis is satisfied. We do have $p^{2}-4q=1$ but we do not have $p^{2}=4q^{2}=(1+2q)^{2}$.

Actually, the difference between the roots is $1$ if and only if $p^{2}-4q=1$. Perhaps there is a typo in the question and the first equality in $p^{2}=4q^{2}=(1+2q)^{2}$ is supposed to be replaced by $+$.
The second one is very likely supposed to be
$$
p^2+4q^2 = (1+2q)^2.
$$
To prove it, try to expand $(1+2q)^2$ and then use the first result. You no longer need to think about the original quadratic for this part.