why I should satisfy these ?
- $e$ should be small.
There's 3 reasons
- It's in the interest of speed of the public-key operations (encryption and signature verification), which in RSA take time roughly proportional to $\log e$.
- It ensures compatibility with some implementations, which limit $e$ to 32 (or is it 31) bits.
- It makes it impossible to use a small $d$, which would ruin security. The question's final paragraph is right about that.
Notice that for mostly historical reasons, you'll also find recommendations that $e$ is not too small. Definitely, $e\le1$ would be a terrible idea. At the end of the day, the most common and unobjectionable is $e=F_4=2^{(2^4)}+1=65537$, where $F_4$ is the largest known Fermat prime.
- $\gcd(\varphi(N),e)=1$.
In this, $\varphi(N)$ differs only in notation from $\phi(N)$ or is it $\Phi(N)$ in the question's "$\phi(N)$ is equal to $(p-1)(q-1)$". That's the Euler totient. $\varphi(N)=(p-1)(q-1)$ holds when $N=p\cdot q$ with $p$ and $q$ distinct primes, which is overwhelmingly likely when $p$ and $q$ are primes chosen randomly enough.
And $\gcd(\varphi(N),e)=1$ is necessary for existence of (integers) $k$ and $d$ such that $e\cdot d=k\cdot\varphi(N)+1$.
$e$ should be an odd number.
Unless $p$ or $q$ is $2$, which would be a bad idea, that follows from $e\cdot d = k\cdot(p-1)(q-1)+1$.
I think the second and third conditions are stated just to ensure that $d$ is an integer too because of $d=(k\cdot\varphi(N)+1)/e$.
Uh, no. Integers and bitstrings are about the only data representations in cryptography, and in RSA they are often assimilated per big-endian convention, so everything is an integer.
That exists $k$ with $e\cdot d=k\cdot\varphi(N)+1$ or (equivalently) $d=(k\cdot\varphi(N)+1)/e$ can be noted $d\equiv e^{-1}\pmod{\varphi(N)}$, and that means $d$ is (an integer representative of) the inverse of $e$ in the multiplicative group of the integers modulo $\varphi(N)$.
