An RSA key with special prime numbers 
$(N,5)$ is a public RSA key with prime numbers $p,q$ such that $N =pq,  q = 2p-3$ and $p>5$. Show that:
  
  
*
  
*$p \equiv 3 \mod 5$
  
*$(N,d)$ with $d =\frac{1}{5} (1+(p-1)(q-1))$ is a valid private RSA key to the public key $(N,5)$.
  

Here is my attempt to proof item 2:
We know that (from the construction of the RSA cryptosystem)
$$d \cdot e ≡ 1 \mod \varphi(N)$$
Since $\varphi(N) = (p-1)(q-1)$ and $e=5$, we have to solve
$$d \cdot 5 ≡ 1 \mod (p-1)(q-1)$$
for $d$ which is the same as 
$$d \cdot 5 ≡ 0 \mod ((p-1)(q-1)+1)$$
and we directly see that
$$d =\frac{1}{5} (1+(p-1)(q-1))$$
which proves item 2.
Is that correct and can anybody help me with item 1?
 A: Item 1. Look at the values $r=p\bmod 5.$ If $r=0$ then $p$ is not prime. If $r=1$ or $r=2$ (i.e. $q\bmod 5=1$) you have $\phi \bmod 5 = 0$ and there is no inverse. $r=4$ implies that $q$ is no prime because $q\bmod 5=0$. 
So if your RSA system is valid, you necessarily have $p\equiv 3 \pmod 5$, and from the definition $q\equiv 3 \pmod 5.$
Item 2.
From 1 we have $\phi = (p-1)(q-1)\equiv 2\times 2\equiv 4 \pmod 5.$ This means that $\phi+1$ is a multiple of $5$. Let $$d = \frac{\phi+1}{5}=\frac{(p-1)(q-1)+1}{5}$$
Then $$5d=5\frac{\phi+1}{5}=\phi+1 \equiv 1 \pmod {\phi},$$
i.e. $d\,$ is the multiplicative inverse of $5$ modulo $\phi$ and a valid private key for $(N,5).$
Note: As already mentioned in a comment, this specific RSA system is insecure, because you can easily factor $N$. You get a quadratic equation in integers from $N=p(2p-3)$
$$2p^2-3p-N=0$$ 
with the positive solution 
$$p=\frac{3+\sqrt{8N+9}}{4}$$
A: *

*For $(N,5)$ to qualify as an RSA public key, we must have $\gcd(\varphi(N),5)=1$.
As $\varphi(N)=(p-1)(q-1)$, this means that $\gcd(p-1,5)=\gcd(q-1,5)=1$, so $p$ and $q$ must equal $0,2,3,$ or $4 \bmod 5$. $p$ and $q$ are prime, and we are told that $p > 5$ and $q=2p-3$, so $p$ and $q$ must be non-zero $\bmod 5$. This leaves $2,3,$ and $4$ as possible values for $p \bmod 5$ and $q \bmod 5$.
Now for each of these possible values for $p \bmod 5$, use $q=2p-3$ to find the corresponding value of $q \bmod 5$. What are the permissible pairs $(p \bmod 5,q\bmod 5)$?

*From $$d \cdot 5 ≡ 0 \mod ((p-1)(q-1)+1)$$ you can't conclude that $$d =\frac{1}{5} (1+(p-1)(q-1))$$
You have to approach it differently:


*

*show (using part 1) that $d=\frac{1}{5} (1+(p-1)(q-1))$ is an integer;

*show that $5d = 1 \bmod \varphi(N)$.


