And we can proof: All solutions of $n$ are $n =1, 3,4,3 \times 5,3 \times 5 \times 17,3 \times 5 \times 17 \times 257,3 \times 5 \times 17 \times 257 \times 65537$.

Can you post a proof for this?

Actually your claim is wrong. n can be any number of the form $F_{0}F_{1}...F{m}$ where F are the Fermat numbers (easy to prove).

NO. These consecutive Fermat numbers must be all prime. Euler showed that $F_5$ is not anymore a prime. His prime factors cannot be Fermat primes, since Fermat numbers are pairwise coprime, and so $\varphi(F_5)$ will not be a power of $2$.

Oh, sorry I knew that but I took incorrectly $\phi({F_{n}})=F_n-1$

Give a complete solution please.

I think I have a solution. It is pretty long, probably the main reason being that I didn't want to make induction. Let $F(m)=2^{2^m}+1$ be the $m$th Fermat number. It is known that $\phi(n)$ is a power of two if and only if $n=F(a_1)F(a_2)\ldots F(a_k)$ (since the statement says $n$ is odd), where $a_1<a_2<\ldots<a_k$. We may suppose $k\geq1$. Similarly we have $n+1=2^\beta F(b_1)F(b_2)\dots F(b_l)$. Assuming $l>0$, the aim is to prove $n=5$. $F(m)$ grows very fast (see e.g. here), so it is relatively easy to study $n$ and $n+1$ modulo $2^e$, for various exponents $e$. With that, we can compare $n$ and $n+1$. Let's begin using modulo $2^{2^{a_1}+1}$: $n\equiv F(a_1) \implies n+1\equiv 2^{2^{a_1}}+2$ To proceed, we split the problem in two cases. Case 1: $a_1>0$. In this case the rhs is even, but not multiple of $4$, whence $\beta=1$. Dividing both sides (and the modulo) by $2$, we get $\frac{n+1}{2}\equiv 2^{2^{a_1}-1}+1\mod 2^{2^{a_1}}$ $\implies F(b_1)F(b_2)\ldots F(b_l)\equiv2^{2^{a_1}-1}+1\mod 2^{2^{a_1}} (*)$ Now, what to do with the lhs? Working modulo $2^{2^{b_1}+1}$, the left side becomes $F(b_1)=2^{2^{b_1}}+1$. We shall argue through the following fact. Lemma: consider nonnegative integers $e_1,f_1,e_2,f_2$ and $x$ such that * The remainder in the division of $x$ by $2^{e_1}$ equals $2^{f_1}$ (in particular $f_1<e_1$). * The remainder in the division of $x$ by $2^{e_2}$ equals $2^{f_2}$ (in particular $f_2<e_2$). Then $f_1=f_2$. Proof: assume $e_1\leq e_2$. Then, in the division by $2^{e_1}$, $2^{f_2}$ leaves the same remainder as $x$, which is $2^{f_1}$. Then $2^{e_1}|2^ {f_2}-2^{f_1}.\square$ Now we choose $e_1=2^{a_1}$ $f_1=2^{a_1}-1$ $e_2=2^{b_1}+1$ $f_2=2^{b_1}$ $x=F(b_1)F(b_2)\ldots F(b_l)-1$ and apply the lemma (see $(*)$), to get $f_1=f_2\implies a_1=1$ and $b_1=0$. Up to now, we have $F(a_1)=5$ and $F(b_1)=3$, so that $n=5F(a_2)\ldots F(a_k)$ $n+1=6F(b_2)\ldots F(b_l)$ For $k=1$, this actually means $n=5$. We're going to arrive at a contradiction supposing $k\geq2$. $n\equiv 5F(a_2)\equiv5.2^{2^{a_2}}+5\mod 2^{2^{a_2}+1}$ $n+1\equiv 2^{2^{a_2}}+6\mod 2^{2^{a_2}+1}$ $\frac{n+1}{2}\equiv 2^{2^{a_2}-1}+3\mod 2^{2^{a_2}}$ $3F(b_2)\ldots F(b_l)\equiv 2^{2^{a_2}-1}+3\mod 2^{2^{a_2}}$ The number $3F(b_2)\ldots F(b_l)$ then yields the remainder $2^{2^{a_2}-1}+3$ after division by $2^{2^{a_2}}$, and we claim it also yields the remainder $3F(b_2)\ldots F(b_j)$, where $b_j<a_2\leq b_{j+1}$. Indeed, $3F(b_2)\ldots F(b_j)$ is a product of distinct Fermat numbers, all smaller than $F(a_2)$, and therefore must be smaller than $F(a_2)$ (see the fourth property at the link above). Thus, $3F(b_2)\ldots F(b_l)= 2^{2^{a_2}-1}+3$ where one number is multiple of $3$ and the other is not. Case 2: $a_1=0$. Equivalently, $F(a_1)=3$. So $n=3F(a_2)\ldots F(a_k)$ $n+1=2^\beta F(b_1)F(b_2)\ldots F(b_l)$ If $n=3$, then $n+1=4$ is a power of $2$, conformly to our desires. Else, we consider the equalities above modulo $2^{2^{a_2}+1}$: $3F(a_2)+1\equiv2^{\beta}F(b_1)F(b_2)\ldots F(b_l)$ $\implies 2^{2^{a_2}}+4\equiv2^{\beta}F(b_1)F(b_2)\ldots F(b_l)$ We again consider two cases. Case 2.1: $a_2>1$. In this case, the lhs is $\equiv 4\mod 8$, so that $\beta=2$. Then $2^{2^{a_2}-2}+1\equiv F(b_1)F(b_2)\ldots F(b_l)\mod 2^{2^{a_2}-1}$ The rhs is congruent to $F(b_1)$ modulo $2^{2^{b_1}+1}$. Applying the lemma once more, $2^{a_2}-2=2^{b_1}\implies a_2=2$ So $F(a_2)=17$, and $n=3\times17\times F(a_3)\ldots F(a_k)$ $n+1=4F(b_1)\ldots F(b_l)$ It is clear that, whenever $m\geq3$, $F(m)\equiv1\mod 256$. Then $n\equiv51\mod 256$, while $n+1$ must be congruent either to $4$ or to $4.5$, depending on whether $F(b_1)=5$ or $F(b_1)\geq 257$. This is a contradiction. Case 2.2: $a_2=1$. We have $F(a_2)=5$, so that $n=3\times5\times F(a_3)\ldots F(a_k)$ $n+1=2^{\beta}F(b_1)\ldots F(b_l)$ If $k=2$, all right, $n=15$ and $n+1=16$. For $k>2$, we easily find the remainder of $n$ modulo $2^{2^{a_3}+1}$, which is $2^{2^{a_3}}+15$ (observe that $2^{2^{a_3}}\geq16$). Case 2.2.1: $a_3>2$, which means $2^{2^{a_3}}>16$. Then $n+1\equiv 2^{2^{a_3}}+16\mod 2^{2^{a_3}+1}$, in turn implying $n+1\equiv 16\mod 32$. Therefore $\beta=4$. This is enough to get $3\times5\times F(a_3)+1\equiv 16F(b_1)\ldots F(b_k)\mod 2^{2^{a_3}+1}$ $2^{2^{a_3}}+16\equiv16F(b_1)\ldots F(b_k)\mod 2^{2^{a_3}+1}$ $2^{2^{a_3}-4}+1\equiv F(b_1)\ldots F(b_k)\mod 2^{2^{a_3}-3}$ Subtracting $1$ and using the lemma yet again (ok, this is getting boring), $2^{b_1}=2^{a_3}-4$, then $a_3=3$. $n=3\times5\times257\times F(a_4)\ldots F(a_k)$ $n+1=16F(b_1)\ldots F(b_l)$ Modulo $2^{2^4}=65536$, we have $n\equiv 3.5.257$. The only other Fermat number below $65536$ is $17$, so $n+1\equiv 16$ or $272\mod 65536$. This is a contradiction. Case 2.2.2: $a_3=2$. Then $n=3\times5\times17\times F(a_4)\ldots F(a_k)$ $n+1=2^{\beta}F(b_1)\ldots F(b_l)$ Here we obtain the solution $n=255$. If $n>255$, let's work modulo $2^{2^{a_4}+1}$: $n\equiv 2^{2^{a_4}}+255$ $n+1\equiv 2^{2^{a_4}}+256$ Case 2.2.2.1: $a_4>3$. Then $n+1\equiv 256\mod 512$, and $\beta=8$. Therefore $n+1\equiv 2^{2^{a_4}}+256\mod 2^{2^{a_4}+1}$ $\frac{n+1}{256}\equiv 2^{2^{a_4}-8}+1\mod 2^{2^{a_4}-7}$ Also, $F(b_1)\ldots F(b_l)\equiv 2^{2^{b_1}}+1\mod 2^{2^{b_1}+1}$ The lemma provides $2^{b_1}=2^{a_4}-8\implies a_4=4$. Then $n=3\times5\times17\times65537\times F(a_5)\ldots\times F(a_k)$ $n+1=256\times F(b_1)\ldots F(b_l)$ But this is incoherent modulo $2^{2^5}$. Case 2.2.2.2: $a_4=3$. $n=3\times5\times17\times257\times F(a_5)\ldots F(a_k)$ If $k=4$, we have the solution $n=65535$. For bigger $n$ we get $3\times5\times17\times257\times F(a_5)\ldots F(a_k)+1\equiv 2^\beta F(b_1)\ldots F(b_l)\mod 2^{2^{a_5}+1}$ $65535(2^{2^{a_5}}+1)+1\equiv 2^\beta F(b_1)\ldots F(b_l)\mod 2^{2^{a_5}+1}$ $2^{2^{a_5}}+65536\equiv 2^\beta F(b_1)\ldots F(b_l)\mod 2^{2^{a_5}+1}$ From this there arise two more cases: Case 2.2.2.2.1: $a_5>4$. Here we find $\beta=16$ and $2^{2^{a_5}-16}+1\equiv F(b_1)\ldots F(b_l)\mod 2^{2^{a_5}-15}$ Then $2^{b_1}=2^{a_5}-16\implies a_5=5$. But $F(5)$ is not a prime! Case 2.2.2.2.2: $a_5=4$. $n=3\times5\times17\times257\times65537\times F(a_6)\ldots F(a_k)$ $n+1=2^\beta F(b_1)\ldots F(b_l)$ As observed above, $F(5)$ is not a prime, therefore $a_6\geq6$. So if we look this modulo $2^{2^{a_6}+1}$ we have $n\equiv(2^{32}-1)F(a_6)$, then $2^{2^{a_6}}+2^{32}\equiv2^\beta F(b_1)\ldots F(b_l)$ This results in $\beta=32$ (it is enough to work modulo $2^{33}$). Finally, $2^{2^{a_6}-32}+1\equiv F(b_1)\ldots F(b_l)\mod 2^{2^{a_6}-31}$ Using the lemma, $2^{a_6}-32=2^{b_1}\implies b_1=5\implies F(b_1)$ is not a prime. The last contradiction we needed!

Any complete solution?

Above is complete solution

Let $F(m)=2^{2^m}+1$ be the $m$th Fermat number. It is known that $\phi(n)$ is a power of two if and only if $n=F(a_1)F(a_2)\ldots F(a_k)$ (since the statement says $n$ is odd), where $a_1<a_2<\ldots<a_k$. We may suppose $k\geq1$. Similarly we have $n+1=2^\beta F(b_1)F(b_2)\dots F(b_l)$ and let's say $ l>0 $ and $ A=\left \{ a_i \right \},B=\left \{ b_j \right \}, A\cap B=\varnothing $ . <Claim> $ \prod_{n=0}^{\infty }\left ( 1+ \frac{1}{2^{2^n}}\right )= 2 $. pf) $ \prod_{n=0}^{\infty }\left ( 1+ \frac{1}{2^{2^n}}\right )=1+\sum_{i_1,...,i_k}^{ }\frac{1}{2^{2^{i_1}+2^{i_2}+...+2^{i_k}}}=1+\sum_{n=1}^{\infty }\frac{1}{2^n}=2 $ Let's say that $ \beta -(\sum 2^{a_i}-\sum 2^{b_j})=X $ We know that $ 2^X\frac{n}{n+1}=\frac{\prod (1+\frac{1}{2^{2^{a_i}}})}{\prod (1+\frac{1}{2^{2^{b_j}}})} $. So, $ \frac{1}{2}<2^X\frac{n}{n+1}<2 $ which means that $ X=0,1 $. Case1) $ \beta >1 $ : $ n \equiv 1(mod2^{2^{a_1}}) $ so, $ a_1=0 $. which means that $ 2^X\frac{n}{n+1}=\frac{\prod (1+\frac{1}{2^{2^{a_i}}})}{\prod (1+\frac{1}{2^{2^{b_j}}})}> \frac{3}{2}\prod_{n=1}^{\infty }(1+\frac{1}{2^{2^n}})^{-1}=\frac{9}{8} $ and so $ X=1 $. So, $ \sum 2^{a_i}-\sum 2^{b_j}=\beta-1>0 $ so, $ max(a_i)=t>max(b_j)=s $ then, $\frac{2n}{n+1}=\frac{\prod (1+\frac{1}{2^{2^{a_i}}})}{\prod (1+\frac{1}{2^{2^{b_j}}})}\leq \prod_{i=0}^{t}(1+\frac{1}{2^{2^{i}}})\times (1+\frac{1}{2^{2^{t-1}}})^{-1}<2(1-\frac{1}{2^{2^{t}}})< 2(1-\frac{1}{n+1}) $ so contradiction Case2) $ \beta =1 $ : $ \beta -(\sum 2^{a_i}-\sum 2^{b_j})=X=0,1 $ so, $ X=0 $. Which means $ \sum 2^{a_i}-\sum 2^{b_j}=1 $ and so $ A=\left \{ t+1 \right \},B=\left \{ 0,1,...,t \right \} $ for some $ t $. And from this it's easy to get $ t=0 $. So, $ n=5 $.

I found really nice solution (at least for me ): $n=(2^{2^{a_1}}+1)...(2^{2^{a_k}}+1), n+1= 2^c(2^{2^{b_1}}+1)...(2^{2^{b_m}}+1)$ where $0<=a_i < a_j , 0<=b_i < b_j$ for all $i<j$ and all $2^{2^{a_i}}+1, 2^{2^{b_i}}+1$ are primes. Notice: $(*)$ if $c>1$ then $3| n$ if $c>2$ then $5|n$ because $n+1 \equiv 3 + 1 = 4 ($mod $2^{2^{a_2}}$) and since $2^3 | n+1$ it must be $2^{a_2}=2$ Similarly: $c>4 \implies 17|n$ $c>8 \implies 257|n$ $c>16 \implies 2^{16} +1|n$ and if $c>32$ then $n+1 \equiv 2^{32} ($mod $2^{2^{a_6}}$) but this is not possible since this implies that $2^{32}+1$ is prime. Notice that here I don't make mistake by assuming there exist next $a_i$ because for example for $c>32$ if it didn't exist $a_6$ then $n=3 \cdot 5 \cdot 17 \cdot 257 \cdot (2^{16}+1) < 2^{32} < 2^c-1<= n+1 -1$ which is clearly a contradiction. Notice that $(**)$: $a_k>b_m$. Since $(n,n+1) =1$ we have $a_k \not = b_m$. Suppose $a_k<b_m$ then: $n<= (2+1)(2^2+1)...(2^{2^{a_k}}+1) = 2^{2^{a_k+1}}-1 <= 2^{2^{b_m}}-1 < n+1 -1$ which gives desired contradiction. Now combine $(*), (**)$ to get: $b_m<4!!$ Suppose $b_m>5$ ( $b_m \not = 5$). Now $n+1 <= 2^{32}(2+1)(2^2+1)(2^{2^3}+1)(2^{2^4}+1)(2^{2^6}+1)...(2^{2^{b_m}}+1) < (2+1)(2^2+1)(2^{2^3}+1)(2^{2^4}+1)(2^{2^5}+1)...(2^{2^{b_m}}+1)=2^{2^{b_m +1}}-1 < 2^{2^{a_k}}+1<=n $ which is clearly a contradiction. If $b_m=4$ we have $n+1 <= 2^{32}(2+1)(2^2+1)(2^{2^3}+1)(2^{2^4}+1)< (2+1)(2^2+1)(2^{2^3}+1)(2^{2^4}+1)(2^{2^5}+1) = 2^{2^6}-1<=2^{2{a_k}}-1<n$ , contradiction. Now we have few cases to check! It might look like there's a lot of work to do but there isn't really. If you use $(*)$ and $(n,n+1)=1$ and make cases for $c<9$ since for $c>8$ you get that $n+1$ is power of 2, and you can easily get what are prime factors of n giving n an upper bound for every case and knowing they must be Fermat's numbers. (I can show more if anyone asks)

I found this problem unexpectedly nice for using "size" as the keyword. First, let me modify the problem just a bit. RaleD wrote: Determine all positive integers $n$ such that both $\phi(n)$ and $\phi (n+1)$ are powers of two, but $n$ is not a power of $2$. I claim the answers are $3, 5, 15, 255, 65535$ and $2^{32} - 1$. Now, let's enjoy the whole ride. Lemma 01. Let $a \in \mathbb{N}$ such that $\varphi(a)$ is a power of $2$, then the prime divisors of $a$ are either $2$ or $2^{2^k} + 1$ for some $k \in \mathbb{N}$. Proof. Notice that if $p | n$, then $p - 1 | \varphi(n)$. Therefore, \[ p - 1 | \varphi(n) = 2^m \]We then have $p = 2^q + 1$ for some $q$. Suppose $q$ has an odd prime divisor, then this results that $p$ is not a prime. Therefore, $q = 2^k$ for some $k \in \mathbb{N}$. Therefore, $p = 2^{2^k} + 1$, or $p = 2$. Lemma 02. There are no multiplicity of odd prime factors in $a$ if $\varphi(a)$ is a power of $2$. Proof. Suppose there exists an odd prime $p$ such that $\nu_p(a) \ge 2$, then $p | \varphi(a) = 2^m$, which is a contradiction. Lemma 03. Consider the largest odd prime dividing either $n$ or $n + 1$, let it be $2^{2^a} + 1$. Then it must divide the odd number. Proof. Suppose not, then the largest odd prime factor divides the even number. By Lemma 02, the odd number will have a maximum value of \[ (2+1)(2^2 + 1)(2^4 + 1)\dots(2^{2^{a-1}} + 1) = 2^{2^a} - 1 \]which is not possible as the even number is at least $2(2^{2^a} + 1)$. Lemma 04. Consider the smallest odd prime number dividing the odd number, let it be $2^{2^b} + 1$. Then either $b = 0$ or $a = b$. Proof. For the sake of contradiction, assume $b \not= 0$. Since it is the smallest prime dividing that odd number, then the other primes must be $1 \ (\text{mod} \ 2^{2^{b + 1}})$, and hence this odd number is $2^{2^b} + 1 \ (\text{mod} \ 2^{2^{b + 1}} )$. Since the even number is only $1$ apart from this odd number, it is either $2^{2^b} + 2 \ (\text{mod} \ 2^{2^{b + 1}} )$ or $2^{2^b} \ (\text{mod} \ 2^{2^{b + 1}})$. Hence, the $\nu_2(\text{even number}) \le 2^b$. But by size argument, since \[ 2^{2^b} + 1 > 2^{2^b} \]\[ 2^{2^a} + 1 > (2^{2^{a - 1}} + 1)(2^{2^{a - 2}} + 1) \dots (2^2 + 1)(2 + 1) > \frac{\text{even number}}{2^{2^b}} \]Therefore, the odd number is strictly greater than 1 plus the even number. Therefore, we mustn't have $a \not= b$. Now, considering $a = b$, then we have the odd number to be $2^{2^b} + 1$, as it is the smallest prime and the largest prime factor that divides itself. Since $n + 1 \not= 2^{2^b} + 1$ by the problem's condition. Then we have $n = 2^{2^b} + 1$. Therefore, \[ n + 1 = 2^{2^b} + 2 = 2(2^{2^b - 1} + 1) \]Since $\varphi(n+1)$ is a power of $2$ as well, by $\textbf{Lemma 01.}$, we have $2^b - 1 = 2^c$ for some $c \in \mathbb{N}$. For parity reason, this is only possible when $b = 1$. This gives us $n = 2^{2^1} + 1 = 5$ as a solution. The problem melts to the case where $b = 0$, that is, the odd number is divisible by 3. If $n + 1$ is odd, since the other prime divisors of the odd number is $1 \ (\text{mod} \ 4)$, then we have \[ n + 1 \equiv 3 \ (\text{mod} \ 4) \Rightarrow n \equiv 2 \ (\text{mod} \ 4) \]Therefore, $\nu_2(n) = 1$. Since $2^{2^a} + 1 > (2^{2^{a - 1}} + 1)(2^{2^{a - 2}} + 1) \dots (2^2 + 1)(2 + 1) $ and $3 > 2$. The odd number is strictly greater than 1 plus the even number, which is not possible, unless $a = 0$ itself. This gives $n = 2$, which is not a solution. We will now consider the case where $n$ is odd, and by a similar reasoning, we have $n \equiv 3 \ (\text{mod} \ 4)$. Lemma 05. If $\nu_2(n+1) \ge 2^{d+1}$, then $n$ is divisible by $2^{2^{d + 1}} - 1$. Proof. Notice that $2^{2^{d+1}} - 1 = (1 + 2)(1 + 2^2)(1+2^4)\dots (1+2^{2^d})$. Suppose not, let $2^k$ be the smallest exponent such that $1+2^{2^k}$ does not divide $n$. Suppose $k \le d$, then \[ (1+2)(1+2^2)\dots(1+2^{2^{k-1}} ) \equiv 2^{2^k} - 1 \ (\text{mod} \ 2^{2^{k+1}} ) \]Since the other prime must be $1 \ (\text{mod} \ 2^{2^{k+1}})$, this gives us $n + 1 \equiv 2^{2^k} \ (\text{mod} \ 2^{2^{k+1}} )$. Hence, $\nu_2(n+1) = 2^k \ge 2^{d+1}$, this contradicts our assumption. Moreover, by $\textbf{Lemma 05.}$, we know that $\nu_2(n+1) = 2^k$ for an integer $k \in\mathbb{N}$. Lemma 06. $n = 2^{2^r} - 1$ for some integer $r \in \mathbb{N}$. Proof. Suppose that the largest prime of $n$, which is the largest odd prime among $n$ and $n + 1$, $2^{2^a} + 1$ is distinct from any prime dividing $2^{2^d} - 1$ by assuming $\nu_2(n+1) = 2^d$. Now, notice that \[ n = (2^{2^a} + 1)(2^{2^d} - 1)N \ge (2^{2^a} + 1)(2^{2^d} - 1) \]\[ n + 1 = 2^{2^d} \cdot M \le 2^{2^d} \cdot \frac{2^{2^a} - 1}{2^{2^d} - 1} \]But, \[ n + 1 \le \frac{2^{2^a} - 1}{2^{2^d} - 1} <(2^{2^a} + 1)(2^{2^d} - 1) \le n \]which is a clear contradiction. Therefore, $2^{2^a} + 1$ must be in the prime dividing $2^{2^d} - 1$, which therefore gives $n = 2^{2^d} - 1$, as there are no more larger prime factors. We just need to check the first few cases as the $2^{32} + 1$ is not a fermat prime. This gives us $2^2 - 1, 2^4 - 1, 2^8 - 1, 2^{16} - 1, 2^{32} - 1$ being the solution as well. Al Fine!

@above nice

im actually stupid me when i write $2^k+1$ for prime factors not realizing they have to be fermat primes (it's trivial if youmake the observation skull) in any case the problem should immediately follow by just expanding out and looking at the exponents of two (since fermat primes, theres no overlap. bruh.) i wasted an hour on this :skull: :skull: basically if you don't make fermat prime observation you literlaly make zero progress bro thats actually so braindead smh

anyway here's the soution to this 0 moh problem (using waltz 01 adaptation on otis) (you find all n not powers of 2 such that phi condition holds) Any odd prime of the form $2^k+1$ must actually be of the form $2^{2^k}+1$. In my defense this isn't intuitively obvious so I completely missed it ugh. The point is that otherwise writing $k=2^e\cdot o$ we find that \[2^{2^e}+1\mid 2^k+1.\] Now write \[n=2^e(2^{2^{k_1}}+1)\dots(2^{2^{k_x}}+1)\]and \[n+1=(2^{2^{\ell_1}}+1)\dots(2^{2^{\ell_y}}+1).\]Expand each of these out; we can actually get $\nu_2(n)=2^{\ell_1}$ from the second equation, hence we can now write \[(2^{2^{k_1}}+1)\dots(2^{2^{k_x}}+1)=\frac{1}{2^{2^{\ell_1}}}((2^{2^{\ell_1}}+1)\dots(2^{2^{\ell_y}}+1)-1).\]The same expansion now yields the sum of a bunch of distinct powers of $2$ on both the left and right-hand sides (since exponents are powers of two. hi binary). Notably this fails if you don't discover that exponents are powers of two (i.e. all Fermat primes). I spent like an hour trying to resolve this to no avail lmfao :skull: Then obviously $2^x=2^y-1$ while $x\ge 1$. This doesn't work. If $n$ is odd then expand again, adding $1$ and dividing by $2^e$ where $\nu_2(n+1)=e$. If all $k_i\ge 1$ then this yields $e=1$. Also $x=y$. At this point the expansion of \[(2^{2^{\ell_1}}+1)\dots(2^{2^{\ell_y}}+1)\]must contain a $2^{2^{k_x}-1}=2^{2^{k_x-1}}\cdot 2^{2^{k_x-2}}\cdot \dots \cdot 2^{2^0}$. In particular we have $k_x\ge x$ hence we must have \[\{\ell_1,\dots,\ell_y\}=\{0,\dots,x-1\}\]and also \[\{k_1,\dots,k_x\}=\{1,\dots,x\}\]To finish this case off just consider the largest exponents in the expansions for $n$ and $n+1$. For $n$ it should be \[2^1+\dots+2^x\]and for $n+1$ it should be \[1+2^0+\dots+2^{x-1}=2^x\]hence $x=1$. As it turns out $n=5$ does work. Yay. Now assume that $k_1=0$. This case is super easy lmao. Basically, add $1$ to expansion of $n$ again. But this time, the expansion will have two of $2^1$. If these combine into a $2^2$ and nothing happens, then the resulting is $2^x-1=2^y$, which would imply $x=1$ and $y=0$. That yields $3$ as a solution. If there is another $2^2$ to combine with, then clearly $k_2=1$. That actually means we can keep combining until we have a lone $2^4$. If nothing combines after this, we would get $2^x-3=2^y$, yielding $y=0$ and $x=2$. That yields $(2^1+1)(2^2+1)=3\cdot 5$ which is fine. If the $2^4$ combines with another $2^4$, then $k_3=2$. Now it can combine up to a $2^8$. If nothing happens from here, then $2^x-7=2^y$, yields $y=0$ and $x=3$. That yields $(2^1+1)(2^2+1)(2^3+1)=3\cdot 5\cdot 17$ which is fine. Keep going until we hit a non-Fermat prime. So solutions are up to $3\cdot 5\cdot 17\cdot 257\cdot 65537$. Done.