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See Vitor's answer for the answer your professor was looking for. However, for any PRNG of the form $s_{i+1} = F(s_{i})$, where the attacker sees the $s_i$ values, and knows $F$, then he can distinguish it. Given a sequence of values $r_1, r_2, ...$, he can determine whether it was generated by that PRNG by checking if $r_2 = F(r_1)$; this is always true ...


$s_i = s_{i-1}\cdot(N + 1) + 1 = s_{i-1} \cdot N + s_{i-1} + 1$ but $s_{i-1} \cdot N = 0 \pmod N$, so $s_i = s_{i-1} + 1 \pmod N$ which means you can discover the next number to be generated just looking to the current one...


No, A is not true. Suppose that $G_1$ is a secure PRG and $G_2(s) = G_1(s) \oplus 1$, obviously $G_2 \neq G_1$ and $G_2$ is a secure PRG. You can see that $G(s) = G_1(s) \oplus G_2(s) = G_1(s) \oplus G_1(s) \oplus 1 = 1$ which is obviously not a secure PRG. Now you have a hint. You should think the rest of the problems.

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