Is it true that $a+bi$ is prime in $\mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $\mathbb{Z}$? Is it true that $a+bi$ is prime in $\mathbb{Z}[i]$ if and only if $a^2+b^2$ is prime in $\mathbb{Z}$?

How can I prove this? Can anybody help me please?

This is not true when $b=0$ and $a$ is an ordinary prime of the form $4k+3$. And for the same reason it is not true if $a=0$ and $b$ is an ordinary prime of the form $4k+3$.
Added: If $a^2+b^2$ is an ordinary prime, then $a+bi$ is a Gaussian prime.  For suppose that $a+bi=(s+ti)(u+vi)$. By a norm calculation we have $a^2+b^2=(s^2+t^2)(u^2+v^2)$.  So since $a^2+b^2$ is prime, one of $s+ti$ or $u+vi$ is a unit.
For the other direction, suppose $a+bi$ is a Gaussian prime, where neither $a$ nor $b$ is equal to $0$. We show that $a^2+b^2$ is an ordinary prime. Proof would be easy if we assume 
standard results that characterize the Gaussian primes. So we try not to use much machinery.
If $a$ and $b$ are both odd, then $a+bi$ is divisible by $1+i$. Then $a+bi$ is an associate of $1+i$, and $a^2+b^2=2$.
So we may assume that $a$ and $b$ have opposite parities. In that case, $a+bi$ and $a-bi$ are relatively prime.  For any common divisor $\delta$ must divide $2a$ and $2b$. Since $a$ and $b$ have opposite parity, any common divisor divides $a$ and $b$, so must have norm $1$ if $a+bi$ are prime. 
Now suppose that  $p$ is a prime that divides $a^2+b^2$. Then $p$ divides $(a+bi)(a-bi)$. 
Note that $p$ cannot be a Gaussian prime, else it would divide one of $a+bi$ or $a-bi$,
Let $\pi$ be a Gaussian prime that divides $p$.  Then $\pi$ divides one of $a+bi$ or $a-bi$. So $\pi$ must be an associate of one of these, and the conjugate of $\pi$ is an associate of the other. Since $a+bi$ and $a-bi$ arer relatively prime, we conclude that $(a+bi)(a-bi)$ divides $p$, which forces $p=a^2+b^2$.
