Note that if $n$ is even then $3\mid2^n+65$ but $3\not | 5^n-3^n$, impossible, so $n$ is odd, and clearly $n=1$ fails. Now we claim all $n>1$ fail. To see why, note that \[-1 = \left( \frac{2^n+65}{5} \right) = \left( \frac{5}{2^n+65} \right) = \left( \frac{5^n}{2^n+65} \right),\] and now if $2^n+65 \mid 5^n-3^n$ this in turn equals \[\left( \frac{3^n}{2^n+65} \right) = \left( \frac{3}{2^n+65} \right) = \left( \frac{2^n+65}{3} \right) = +1,\] a contradiction. (Here we make use Jacobi symbols.)

1998 n5 Suppose $2^n+65\mid5^n-3^n$. Since $3\nmid 5^n-3^n$ and $5\nmid 5^n-3^n$, we get $n\equiv1\pmod2$. Therefore, $5^n-3^n=5x^2-3y^2$ for some $x$ and $y$, implying $15\equiv\frac{(3y)^2}{x^2}\pmod{2^n+65}$, so $\left(\frac{15}{2^n+65}\right)=1$. Clearly, $2^1+65\nmid5^1-3^1$, so $n\geq2$. However, we also have \begin{align*} \left(\frac{15}{2^n+65}\right)&=\left(\frac{2^n+65}{15}\right)\\ &=\left(\frac{2^n+65}5\right)\left(\frac{2^n+65}3\right)\\ &=\left(\frac{2^n}5\right)\left(\frac{2^n+2}3\right)\\ &=(-1)(1)\\ &=-1, \end{align*}contradiction. Therefore, $2^n+65\nmid 5^n-3^n$ for all $n$.

Quite similar to Romania TST 2008/3/3 and exactly the same approach works here. Definitely easier than N5 and N6.

Isn't the wording supposed to be this one? Quote: Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$.

No such $n$. By taking modulo $3$, $n$ is odd, and it is clear that $n=1$ fails, so assume $n \geq 2$. Therefore, if $(\tfrac{5}{3})^n-1 \equiv 0 \pmod{2^n+65}$, $\tfrac{5}{3}$ must be a quadratic residue modulo $2^n+65$. However, letting $(\tfrac{a}{b})$ denote the Jacobi symbol, by quadratic reciprocity we have $$\left(\frac{5/3}{2^n+65}\right)=\left(\frac{5}{2^n+65}\right)\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{5}\right)\left(\frac{2^n+65}{3}\right)=\left(\frac{\pm 2}{5}\right)\left(\frac{1}{3}\right)=(-1)(1)=-1,$$contradiction. $\blacksquare$ Remark: "IT'S [redacted] JACOBI" [dies]

Is this really n8? It's just some simple quadratic residue stuff

Definition: In the solution below, $\left( \frac{a}{b} \right)$ denotes the Jacobi symbol. No solutions. Since $n = 1$ fails, assume that $n > 1$. Note that $2^n + 65 \equiv 1\pmod 4$. Clearly $n$ even fails, as $3$ would divide $2^n + 65$ but $3$ cannot divide $5^n - 3^n$. Now assume $n$ is odd. Notice that $\frac{\left( \frac{5^n}{2^n + 65} \right)}{ \left( \frac{3^n}{2^n + 65} \right) }$ is equal to $1$ (since $5^n\equiv 3^n \pmod{2^n + 65}$), since $n$ is odd, this implies $\frac{\left( \frac{5}{2^n + 65} \right)}{ \left( \frac{3}{2^n + 65} \right) } = 1$. However, we have that \[\left( \frac{5}{2^n + 65} \right) = \left( \frac{2^n + 65}{5} \right) = -1\]and \[\left( \frac{3}{2^n + 65} \right) = \left( \frac{2^n + 65}{3} \right) = \left( \frac{1}{3} \right) = 1,\]absurd. Note: Another way to show that $\frac{\left( \frac{5}{2^n + 65} \right)}{ \left( \frac{3}{2^n + 65} \right) } = 1$ is to let $p$ be a prime dividing $2^n + 65$. Then $1 = \left( \frac{ \frac{5}{3} }{p} \right) = \frac{ \left( \frac{5}{p} \right) }{ \left( \frac{3}{p} \right)} $. Since this is true for all primes $p\mid 2^n + 65$, we have the desired result.

Why are N8s easier than N5s? I claim the answer is $n=0$. Assume for contradiction that there exists a solution for $n > 0$. Taking $\pmod{3}$ of the fractional expression, it is clear that $n$ must be odd (because otherwise $3$ divides the denominator but not the numerator). Let $(\tfrac{a}{b})$ denote the Jacobi symbol henceforth. Note that $(5/3)^n \equiv 1 \pmod{2^n+65}$, and as $n$ is odd, it is clear that $5/3$ must be a quadratic residue (mod $2^n+65$). But: \begin{align*} \left(\frac{5/3}{2^n+65}\right) &= \left(\frac{5}{2^n+65}\right) \left(\frac{3^{-1}}{2^n+65}\right) \\ &= \left(\frac{5}{2^n+65}\right) \left(\frac{3}{2^n+65}\right) &= \left(\frac{2^n+65}{3}\right) \left(\frac{2^n+65}{5}\right) &= 1 \cdot \left(\frac{2}{5}\right)^n &= 1 \cdot (-1) &= -1, \end{align*}a contradiction, and we conclude. $\blacksquare$

lol what Assume FTSOC that there does exist some $n$ such that $2^n+65\mid 5^n-3^n$. If $n$ is even, then $2^n+65$ is a multiple of $3$, impossible. So $n$ is odd. Let $p$ be a prime dividing $2^n+65$. Obviously $p\not\in\{2,3,5\}$. Then \[p\mid 2^n+65\mid 5^n-3^n\mid 15^n-9^n,\]so $15$ is a QR mod $p$. By Quadratic Reciprocity, \[\left(\frac{p}{3}\right)\left(\frac{p}{5}\right)=\left(\frac{15}{p}\right)\left(\frac{p}{15}\right)=(-1)^{(p-1)/2}.\]Caseworking, we get that $p$ mod $60$ is one of $S=\{1,7,11,17,43,49,53,59\}$. Under further inspection, we note that $S$ is closed under multiplication mod $60$, so $2^n+65$ mod $60$ is also in $S$. So $2^n$ mod $60$ is in \[\{2,6,12,38,44,48,54,56\}.\]Since it must be a multiple of $4$ and not a multiple of $3$, it must be one of $\{44,56\}$. But that means $2^n$ is $1$ or $4$ mod $5$, impossible since $n$ is odd. $\blacksquare$

May I give a small comment to the placement of this problem? It seems that this problem is a kind of a problem that gets killed with one trick... Which reminds me of ISL 2017 N8. The difficulty in finding the trick is supposed to be what makes the problem an N8. The difference between this thing and ISL 2017 N8, in my opinion, is that the kind of trick that kills ISL 2017 N8 is something that might be too hard to find in a contest setting. Meanwhile this? I'm not so sure about that... Anyways, let's replace $5$ and $65$ with arbitrary odd positive integers, say $A$ and $B$, and see how well this problem generalizes. For $n$ even, you want $3 \mid 2^n + B$, so $B \equiv 2 \pmod{3}$; also you want $3 \nmid A$. For $n = 1$, you just want $B + 2 \nmid A - 3$. For $n$ odd with $n > 1$, applying Jacobi symbol, you want that $$ \left(\frac{A}{2^n + B}\right) = -\left(\frac{3}{2^n + B}\right) = (-1)^{(B + 1)/2} \left(\frac{2^n + B}{3}\right) = (-1)^{(B + 1)/2}. $$For a huge convenience, you also want $A \mid B$, so $$ \left(\frac{A}{2^n + B}\right) = (-1)^{(A - 1)(B - 1)/4} \left(\frac{2}{A}\right). $$We can now pick $B \equiv 1 \pmod{4}$ and $A \equiv 3, 5 \pmod{8}$ (as in the original problem). In summary... Quote: If $A \equiv 3, 5 \pmod{8}$, $3 \nmid A$, $B \equiv 5 \pmod{12}$, $A \mid B$, and $B + 2 \nmid A - 3$, then $2^n + B$ does not divide $A^n - 3^n$ for any $n \geq 1$. Quote: If $A \equiv 3, 5 \pmod{8}$, $3 \nmid A$, $B \equiv 5 \pmod{12}$, and $A \mid B$, then $2^n + B$ does not divide $A^n - 3^n$ for any $n > 1$.

Huh. Assume that there exists an $n$ such that $\frac{5^n-3^n}{2^n+65}$ is an integer. By taking mod $3$, $n$ must be odd. Clearly, $n=1$ fails. Let $\left (\frac{a}{b}\right)$ denote the Jacobi symbol. We then have that $$\frac{\left(\frac{5^n}{2^n+65}\right)}{\left(\frac{3^n}{2^n+65}\right)}=\frac{\left(\frac{5}{2^n+65}\right)}{\left(\frac{3}{2^n+65}\right)}=1.$$ But $$\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{3}\right) = 1$$and $$\left(\frac{5}{2^n+65}\right) = \left(\frac{2^n+65}{5}\right) = -1,$$ a contradiction.

Is this really difficult enough to be N8?

We use the Jacobi symbol. Taking $\pmod 3$ forces $n$ odd. We have for $n\ge 3$ \[\left(\frac{5\cdot 3^{-1}}{2^n+65}\right)=\left(\frac{5}{2^n+65}\right)\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{5}\right)\left(\frac{2^n+65}{3}\right)=(-1)(1)=-1\]a contradiction because we need $5\cdot 3^{-1}\equiv 1\pmod {2^n+65}$.

Here is the official answer proposed by IMO committee.

I've just started pursuing math at this level, so I'm not at all good at all of this. With that being said, does anyone know if there is another way to solve this without the use of modules?

Let's assume $n$ is even, then $3 \mid 2^n + 65$ but $3 \nmid 5^n - 3^n$ contradiction. Since $n \neq 1$ we get $n \ge 3$ odd number. Now let's assume that $2^n + 65 \mid 5^n-3^n$ then we get $5^n \equiv 3^n ($mod $ 2^n+65)$ Now let us introduce the jaccobi symbol. We get $$\left( \frac{5^n}{2^n+65} \right) = \left (\frac{3^n}{2^n+65} \right)$$ Now lets count them seperatly: $$\left( \frac{3^n}{2^n+65} \right) = \prod_{i=1}^n \left( \frac{3}{2^n+65} \right)^n = \left( \frac{3}{2^n+65} \right) $$ But since $2^n + 65 \equiv 1 ($mod $ 4) $ we get $\left( \frac{3}{2^n+65} \right) = \left( \frac{2^n+65}{3} \right) = 1 $ $$\left( \frac{5^n}{2^n+65} \right) = \prod_{i=1}^n \left( \frac{5}{2^n+65} \right)^n = \left( \frac{5}{2^n+65} \right) $$ Since $5 \equiv 1 ($mod $ 4)$ we get $\left( \frac{5}{2^n+65} \right) = \left( \frac{2^n+65}{5} \right)$ But $2^n \equiv 2;3 ($mod $ 5)$ and $5 \mid 65$ so $\left( \frac{2^n+65}{5} \right) = -1$ a contradiction

tastymath75025 wrote: Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$. $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Here $\left(\frac{a}{b}\right)$ is the Jacobi symbol Suppose there exists $n$ such that $$2^n+65|5^n-3^n$$If $n=1\Rightarrow 67|2(\Rightarrow \Leftarrow)$ $$\Rightarrow n>1$$$\color{red}\boxed{\textbf{If n is even:}}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow 2^n\equiv 1\pmod{3}$$$$\Rightarrow 2^n+65\equiv 0\pmod{3}$$$$\Rightarrow 3|5^n-3^n$$$$\Rightarrow 3|5^n(\Rightarrow \Leftarrow)$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red}\boxed{\textbf{If n is odd:}}$ $\color{red}\rule{24cm}{0.3pt}$ $$2^n+65|5^n-3^n$$$$\Rightarrow \left( \frac{5^n}{2^n+65} \right)=\left( \frac{3^n}{2^n+65} \right)$$$$\Rightarrow \left( \frac{5}{2^n+65} \right)^n=\left( \frac{3}{2^n+65} \right)^n$$$$\Rightarrow \left( \frac{5}{2^n+65} \right)=\left( \frac{3}{2^n+65} \right)$$$$\Rightarrow \left( \frac{2^n+65}{5} \right)(-1)^{\frac{(2^n+64)(4)}{4}}=\left( \frac{2^n+65}{3} \right)(-1)^{\frac{(2^n+64)(2)}{4}}$$Since $n>1\Rightarrow \frac{(2^n+64)(2)}{4}$ is even: $$\Rightarrow \left( \frac{2^n}{5} \right)=\left( \frac{2^n-1}{3} \right)$$$$\Rightarrow \left( \frac{2(2^{n-1})}{5} \right)=\left( \frac{1}{3} \right)=1$$Since $n$ is odd $\Rightarrow 2^{n-1}$ is a perfect square $$\Rightarrow \left( \frac{2}{5} \right)=1$$$$\Rightarrow -1=1(\Rightarrow \Leftarrow)$$$\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow 2^n+65\nmid 5^n-3^n$$$\color{blue}\rule{24cm}{0.3pt}$

pretty same as #2. n is odd . Let k= 2^n +65 then we have 5^n = 3^n mod k then (15/k)=1 this means (3/k)= (5/k) but we have (3/k) = 1 and (5/k) = -1 which is contradiction.

Jacobi kills it, probably same as others. ISL 2022 N8 wrote: Prove that $2^n+65$ does not divide $5^n-3^n$ for any positive integer $n$ Clearly $n$ has to be odd, because if it is even then $3 \mid 2^n+65 \mid 5^n-3^n$ which is a contradiction, $5^n \equiv 3^n (\mod 2^n+65)$ then: $\left(\frac{5^n}{2^n+65}\right)=\left(\frac{5}{2^n+65}\right)=\left(\frac{2^n+65}{5}\right)=-1$ $\left(\frac{3^n}{2^n+65}\right)=\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{3}\right)=1$ where the last 2 lines are in jacobi symbol. Contradiction...

Assume $2^n + 65 \mid 5^n - 3^n$. If $n \equiv 0 (2)$, then $3 \mid 2^n + 65 \mid 5^n - 3^n$, a contradiction. Thus $n \equiv 1 (2)$. Now note that $\left(\frac{5^n}{2^n + 65} \right) = \left(\frac{3^n}{2^n + 65}\right)$. By using quadratic reciprocity law, we get $\left(\frac{3^n}{2^n + 65}\right) = \left(\frac{2^n + 65}{3^n}\right) = \left(\frac{2^n + 65}{3}\right)^n = \left(\frac{2^n + 65}{3}\right) = \left(\frac{1}{3}\right) = 1$. But $\left(\frac{5^n}{2^n + 65}\right) = \left(\frac{2^n + 65}{5^n}\right) = \left(\frac{2^n + 65}{5}\right) = \left(\frac{2^n}{5}\right) = \left(\frac{2}{5}\right) = -1$, which is an evident contradiction. $\blacksquare$

Assume some $n$ works. Obviously $n=1$ does not work for size reasons. And also even $n$ does not work because we get $3 \mid 2^n+65 \mid 5^n-3^n \implies 3 \mid 5$, which I think is wrong as far as I know. See that we get \[\left(\frac{5^n}{2^n+65}\right)=\left(\frac{3^n}{2^n+65}\right) \iff \left(\frac{5}{2^n+65}\right)=\left(\frac{3}{2^n+65}\right) \iff \left(\frac{2^n+65}5\right)=\left(\frac{2^n+65}3 \right) \iff \left(\frac{2^n}5 \right)=\left(\frac{2^n-1}3\right)\]which is obviously false for odd $n$ because it implies $\left(\frac{2^n}5\right)=-1$ but $\left(\frac{2^n-1}3\right)=1$, a contradiction.

First $n\ne 1$ clearly and $n\not\equiv 0\pmod 2$ by $\pmod 3.$ Now if $p\mid 5^n-3^n$ then $(5/3)^n\equiv 1\pmod n$ with $n$ odd, so $\left(\frac{5/3}p\right)=\left(\frac{15}p\right)=1.$ Now we must have $\left(\frac{15}{2^n+65}\right)=1,$ which is equivalent by QR to $\left(\frac{2^n+5}{15}\right)=1.$ But $2^n\equiv 2,8\pmod{15}$ for odd $n,$ and we can check $\left(\frac7{15}\right)=\left(\frac{13}{15}\right)=-1,$ contradiction.

What's the motivation behind using the Jacobi symbol here?

Can anyone solve it without Jacobi symbol?

Clearly when $n$ is even we get a contradiction, so $n$ is odd and $n$ $\geq$ 3. Now, using Jacobi's symbol we get $\left(\frac{5^n}{2^n+65}\right)=\left(\frac{5}{2^n+65}\right)=\left(\frac{2^n+65}{5}\right)=-1$, and $\left(\frac{3^n}{2^n+65}\right)=\left(\frac{3}{2^n+65}\right)=\left(\frac{2^n+65}{3}\right)=1$. But this two must be equal, then we have a contradiction.

Solved on the recommendation of the orz Sammy27. If $n$ is even then $A=2^n +65$ is divisible by $3$ which doesn't divide $5^n-3^n$. Henceforth assume $n$ is odd. This implies $5^n \equiv 3^n(mod \ A) \implies \left( \frac{5^n}{A} \right) =\left( \frac{3^n}{A} \right)$. Now since the jacobi function is multiplicative, we have $\left( \frac{5^n}{A} \right) = \left( \frac{5}{A} \right)= \left( \frac{A}{5} \right)=-1$ And $\left( \frac{3^n}{A} \right) = \left( \frac{3}{A} \right)= \left( \frac{A}{3} \right)=1$ Which is a contradiction, hence there's no solutions for any $n$.

any sols without using jacobi?

Claim: $n$ is odd Note that $3\nmid 5^n-3^n$, hence $3 \nmid 2^n + 65$. As: \[\left\{\begin{array}{cl} 2^n \equiv 1&n \mbox{ odd} \\ 2^n \equiv 2&n \mbox{ even} \end{array}\right.\]$2^n+65\equiv 2^n+2 \equiv \{0,1\} \pmod{3}$ and for $2^n+65\equiv 1\pmod{3}$, we must have $n$ odd. Therefore, looking at $\pmod{5}$, we have: \[\left\{\begin{array}{cc} 2^n\equiv2&n\equiv1\pmod{4} \\ 2^n\equiv4&n\equiv2\pmod{4} \\ 2^n\equiv3&n\equiv3\pmod{4} \\ 2^n\equiv1&n\equiv4\pmod{4} \\ \end{array}\right.\]and as $n$ is odd, $2^n \equiv 2,3 \pmod{5}$ We now rewrite $(\frac{5}{3})^n\equiv1 \pmod{2^n+65}$, consequently $(\frac{5}{3})^{n+1}\equiv \frac{5}{3} \pmod{2^n+65}$, implying $\frac{5}{3}$ is $QR$ $\pmod{2^n + 65}$. Nonetheless:\[\left( \frac{(\frac{5}{3})}{2^n +65}\right) \equiv \left( \frac{5}{3}\right)\left( \frac{5}{2^n+65}\right)\equiv \left( \frac{2^n+65}{5}\right)\equiv\left( \frac{\{2,3\}}{5}\right)\]but: \[\left( \frac{2}{5}\right), \left( \frac{3}{5}\right)\equiv -1\]therefore as: \[\left( \frac{(\frac{5}{3})}{2^n +65})\right)=\prod_{p\mid 2^n +65}\left(\frac{(\frac{5}{3})}{p}\right)\]there exists $p\mid 2^n +65$ prime such that $(\frac{5}{3})$ isn't $QR$ $\pmod{p}$, contradicting the fact that $(\frac{5}{3})$ is $QR$ $\pmod{2^n +65}$