Don't you just use difference of 2 squares.? Bomb

Let n^2+3^n be = to a^2. Then, by difference of squares, we get: 2n=3^x-3^y where x+y=n. For n>3, we prove by induction that RHS>LHS, contradiction. So checkingg yields n=1 and 3. Bomb

Sorry, this probably sounds really stupid but how do you get from $n^{2}+3^{n}=a^{2}$ to $2n=3^{x}-3^{y}$, $x+y=n$? hope you could clarify. Thanks

RHS>LHS What it means

JL wrote: Sorry, this probably sounds really stupid but how do you get from $n^{2}+3^{n}=a^{2}$ to $2n=3^{x}-3^{y}$, $x+y=n$? hope you could clarify. Thanks I guess he means $(a+n)(a-n)=3^n=3^x\cdot 3^y$ (letting $n=x+y$ of course). Then if say $x\geq y$ we must have $a+n=3^x$ , $a-n=3^y$ . So both of them gives us $2n=3^x-3^y$ .

Hi shyong,do you now what means RHS>LHS :

I'm sure shyong shall forgive me, whether it's me to reply your question, tiks! Well, it means the "right hand side" (RHS) is greater than the left hand one (LFS), sic et simpliciter.

thanks shyong!

Tiks wrote: Hi shyong,do you now what means RHS>LHS : I think he means for $n>3$ , $2n<3^x-3^y$ is always true . since we know that $min\{3^x-3^y\}=3^{p+1}-3^p=3^p\cdot 2$ where $2p+1=n$ But bernoulii inequality tell us that for $p\geq 1$ $3^p=(1+2)^p\geq 1+2p = n$ the equality holds when $p=1$ which is $n=3$ . So for any $n>3$ we always have $3^x-3^y>2n$ , and there is no solution for the equality to hold for $n>3$ .

Afterwards that?

Valentin Vornicu wrote: Find all positive integers $n\geq 1$ such that $n^2+3^n$ is the square of an integer. Bulgaria Let $n^2 + 3^n = x^2$,we put $n^2$ at right and get $3^n = (x - n)(x+n)$. Let $x - n = 3^y$ and $x + n = 3^{n-y}$,of course $n - y > y$,$2y< n$.$x = 3^{n-y} - n$ and we get that $n^2 + 3^n = (3^{n-y} - n)^2$ or $3^n + 2n3^{n-y} = 3^{2(n-y)}$ or $2n3^{n-y} = 3^n(3^{n-2y}-1)$,since $3^{n-2y} - 1 = 2(mod 3)$ we obtain $n = 3^y$ and $3^{n-2y} - 1 = 2$ we solve the system and obtain the equality $n = 3^y = 2y + 1$ $=>$ $y$ can be 0 or 1 and we finaly get $n = 1$ or $n = 3$.

neverlose wrote: Valentin Vornicu wrote: Find all positive integers $n\geq 1$ such that $n^2+3^n$ is the square of an integer. Bulgaria Let $n^2 + 3^n = x^2$,we put $n^2$ at right and get $3^n = (x - n)(x+n)$. Let $x - n = 3^y$ and $x + n = 3^{n-y}$,of course $n - y > y$,$2y< n$.$x = 3^{n-y} - n$ and we get that $n^2 + 3^n = (3^{n-y} - n)^2$ or $3^n + 2n3^{n-y} = 3^{2(n-y)}$ or $2n3^{n-y} = 3^n(3^{n-2y}-1)$,since $3^{n-2y} - 1 = 2(mod 3)$ we obtain $n = 3^y$ and $3^{n-2y} - 1 = 2$ we solve the system and obtain the equality $n = 3^y = 2y + 1$ $=>$ $y$ can be 0 or 1 and we finaly get $n = 1$ or $n = 3$. Sorry for revival, but how do you go from $3^{n-2y}-1\equiv 2\mod3$ to $n=3^y?$

@shyong why n must be odd??

Solution: Let $3^n+n^2=k^2$, where $k$ is a positive integer. So, $3^n=k^2-n^2$. Let $d=\gcd(k,n). \Rightarrow d|3^n$. Let $d=3^e$. Let $n=3^e \cdot n', k=3^e \cdot k'$, where $\gcd(k',n')=1$. The equation becomes $3^{n-2e} = (k'-n')(k'+n')$. If $k'-n'>1$, then clearly $k'+n'>1$, which will imply that both $k'$ and $n'$ are divisible by $3$, which contradicts the fact that $k'$ and $n'$ are co-prime. So, $k'-n'=1. \Rightarrow k'=n'+1$. So, $2n'+1=3^{3^e\cdot n' - 2e}$. We can notice that RHS grows very fast compared to LHS. Easy bounding gives us $n'=1$. So, $3=3^{3^e-2e}$. So, $1=3^e-2e. \Rightarrow 2e+1=3^e. \Rightarrow e=0,1.$ So, $n=1\cdot 3^0, 1\cdot 3^1 = 1,3. \blacksquare$

After rearranging we get: $(k-n)(k+n) = 3^n$ Let $k-n = 3^a, k+n = 3^{n-a}$ we get: $2n = 3^a(3^{n-2a} - 1)$ or, $(2n/(3^a)) + 1 = 3^{n-2a}$ Now, it is clear from above that $3^a$ divides $n$. so, $n \geq 3^a$ If $n = 3^a, n - 2a = 3^a - 2a \geq 1$ so $RHS \geq 3$ But $LHS = 3$ If $n > 3^a$ then $RHS$ increases exponentially compared to $LHS$ so $n$ cannot be $> 3^a$. Thus $n = 3^a$. Substituting value of $n$ above we get: $3 = 3^{3^a - 2a}$ or $3^a - 2a = 1$ this results in only $a = 0$ or $a = 1$ Thus $n = 1$ or $3$.

Here's my work. Find all positive integers $n \geq 1$ such that $n^2+3^n$ is the square of an integer. Solution - Note that $$n^2+3^n=k^2 \implies k^2-n^2=3^n \implies (k+n)(k-n)=3^n.$$Let us experiment a bit with the problem. We can create a table showing values that work when $n=1,2,3,...,9.$ (We are only going to $9$ for a specific reason. Read until the end to find out.) We seek a pattern in the values that work. Here's the table: $$n=1; (k+1)(k-1)=3^1 \rightarrow k=2$$$$n=2; (k+2)(k-2)=3^2 \rightarrow k \notin \mathbb{R}$$$$n=3; (k+3)(k-3)=3^3 \rightarrow k=6$$$$n=4; (k+1)(k-1)=3^4 \rightarrow k \notin \mathbb{R}$$$$n=5; (k+1)(k-1)=3^5 \rightarrow k \notin \mathbb{R}$$$$n=6; (k+1)(k-1)=3^6 \rightarrow k \notin \mathbb{R}$$$$n=7; (k+1)(k-1)=3^7 \rightarrow k \notin \mathbb{R}$$$$n=8; (k+1)(k-1)=3^8 \rightarrow k \notin \mathbb{R}$$$$n=9; (k+1)(k-1)=3^9 \rightarrow k \notin \mathbb{R}.$$Consider the following claim. Claim. $n=1,3$ are the only solutions. Proof. Notice that these solutions claimed are in the form $3^k,$ where $k \in {\{0,1\}}.$ It suffices to show that $k>3$ is, indeed impossible. (Then, all we have to do is show that $n=2$ does not work, and we are done.) Applying Bernoulli's Inequality, we obtain $$2n<3^x-3^y ~ \text{(As shown in Post 10)},$$and the result follows. We now only use casework on $n=1,2,3$ to get that $n=1,3$ work, but $n=2$ does not. All cases have been exhausted, and thus the proof is complete. $~ \blacksquare$ Remark. The table was nothing but motivation for the claim.