If $p$ is a prime dividing $2^{2^k}+1$, then $2$ has order $2^{k+1}$ $\mod p$. This helps us finding the following list: We have primes $p_1=3, p_2=5,p_3=17,p_4=257,p_5=65537,p_6=641,p_7=6700417$ such that the order of $2$ modulo those primes is $2^1 , 2^2, 2^3, 2^4 , 2^5 , 2^6 , 2^6$ respectively. We now can write $\mathbb{N}_0$ as the union of arithmetic sequences $A_1, A_2, ... , A_6 , A_7$ with differences $2^1 , 2^2, 2^3, 2^4 , 2^5 , 2^6 , 2^6$ (e.g.: the sequences of numbers $\equiv 2^{i-1} \mod 2^i$ for $i=1,2,..,6$ together with those $\equiv 0 \mod 2^6$) and call the starting terms $a_1, a_2, ... , a_6 , a_7$ and the differences $b_1, b_2, ... , b_7$ respectively (so we could take $a_i=2^{i-1}, b_i=2^i$ for $i=1,2,...,6$ and $a_7=0, b_7=2^6$). We then choose an integer $k$ by the chinese remainder theorem such that $k \cdot 2^{a_i} \equiv -1 \mod p_i$ . Then for each $n \in \mathbb{N}_0$, $n$ is contained in one of the sequences $A_i$, and additionally $2^n \mod p_i$ only depends on $n \mod b_i$, yielding $k\cdot 2^n+1 \equiv k \cdot 2^{a_i}+1\equiv 0 \mod p_i$. Thus if we've chosen $k$ big enough, too (which we can), none of the numbers$ k\cdot 2^n+1$ is prime, each is divisible by one (and exactly one) of the primes listed above.

if $ n\equiv 0 \pmod2$ then we can choose $ k = 4$, otherwise $ k = 2$, in fact we have the form $ a^4 + 4b^4$ that is composite. t0rajir0u yes, i've read too fast the text, you're rigth, scuse me

The problem asks for a single integer $ k$ that works for all values of $ n$.

Let $F_n = 2^{2^n} + 1$ for each $n \in \mathbb{N}$. We claim that if $F_n$ is composite for some $n \in \mathbb{N}$, then the assertion in the problem is true. Assume that $m$ is the smallest positive integer satisfying that $F_m$ is composite and let $F_m = a \cdot b$. Now note that $F_0, F_1, \dots, F_{m-1}$ are all distinct prime numbers. Furthermore, note that \[F_0 \cdot F_1 \cdots F_{m-1} = (2-1)(2+1)(2^2 + 1)\cdots (2^{2^{m-1}}+1)=2^{2^m}-1=F_m - 2\] which implies that $F_0, F_1, \dots, F_{m-1}$ are coprime to $F_m$ and therefore to $a$ and $b$. Now consider a $k \in \mathbb{N}$ that satisfies \[k \cdot 2^{(2^{i})-1} \equiv -1 \pmod{F_i}\] for all $0 \le i \le m-1$, and which satisfies \[k \cdot 2^{(2^{m})-1} \equiv -1 \pmod{a},\] \[k \equiv -2 \pmod{b}.\] Note that such a $k$ exists by the chinese remainder theorem. Now we will prove that $k\cdot 2^n +1$ is divisible by one of $F_0, F_1, \dots, F_{m-1}, a, b$ for all $n \in \mathbb{N}_0$. Suppose that the largest power of $2$ dividing $n+1$ be $i$ and let $n=2^{i} (2q+1)-1$ where $q \in \mathbb{N}$. Now if $0 \le i \le m-1$, then since $F_i | 2^{2^{i+1}}-1$, \[k\cdot 2^n +1 \equiv k\cdot (2^{2^{i+1}})^q \cdot 2^{2^{i}-1} +1 \equiv k \cdot 2^{2^{i}-1} +1 \equiv 0 \pmod{F_i}.\] If $i=m$, then \[k\cdot 2^n +1 \equiv k\cdot (2^{2^{m+1}})^q \cdot 2^{2^{m}-1} +1 \equiv k \cdot 2^{2^{m}-1} +1 \equiv 0 \pmod{a}.\] If $i \ge m+1$, then let $n=2^{m+1}\cdot r-1$ where $r \in \mathbb{N}$. Now, since $b| F_m | 2^{2^{m+1}}-1$, \[k\cdot 2^n +1 \equiv (-2) \cdot 2^{2^{m+1} \cdot r -1} +1 \equiv -(2^{2^{m+1}})^r +1 \equiv 0 \pmod{b}.\] Hence $k\cdot 2^n +1$ is divisible by one of $F_0, F_1, \dots, F_{m-1}, a, b$ for all $n \in \mathbb{N}_0$. Let $2^l$ be such that $l \in \mathbb{N}$ and $2^l > F_0, F_1, \dots, F_{m-1}, a, b$. As proven previously, $(k\cdot 2^l) \cdot 2^{n} +1 = k \cdot 2^{l+n}+1$ is divisible by one of $F_0, F_1, \dots, F_{m-1}, a, b$ for all $n \in \mathbb{N}_0$. Since $(k\cdot 2^l) \cdot 2^{n} +1> 2^l > F_0, F_1, \dots, F_{m-1}, a, b$, it follows that $(k\cdot 2^l)\cdot 2^n +1$ is composite for all $n\in \mathbb{N}_0$, thus proving the claim. Since Euler proved that $F_5$ is composite, the claim yields the desired result.

The topic is also covered by R.K. Guy in his Unsolved Problems in Number Theory, B21.

SnowEverywhere wrote: Since Fermat proved that $F_5$ is composite Euler did, Fermat's work was mainly making conjectures and completely unbased claims of having a proof.

1982 USAMO #4

Here is a sketch of a bit easier proof, (still not really easy) Looking k in mods 3,5,7,13,19,37 and 73 , So for n , You can eliminate 1 mod from 2 (because we choose k in mod 3) , 1 mod from 4 (k in mod 5) ,1 mod from 3( k in mod 7 realise 2 is quadratic) ...... going on like this you can eliminate all mods from 36 , ( You need to see 2 is quadratic for 2,19 and 73.) For example we look in 36 instead of 72 for prime 73. So for all n's , The expression divides one of 3,5,7,13,19,37 or 73. Since we choose appropriate mod for k in 3.5.7.....73 there are infinetly many of k's ,so we choose k>73. And now n is divisible by one of them and it can't be equal to them so n is not prime for all integers greater than 0.