Hmm...so $3^a+3^b+3^c=x^2$

Yes, it is easy to solve so it was the first question

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taking modulo 8 gives that $a,b,c$ are all odd. let $a \leq b \leq c$ so $3^a(3^{c-a}+3^{b-a}+1 )=x^2$ but $v_3(x^2)=a$ which is odd unless $a=b=c$ which is a solution.

In other words: Choose three numbers (repeats allowed) from the set $\{3,9,27,81...\}$ and add them such that you get a number that falls under the form $x^2$ such that $x$ is a positive integer.

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Yaghi wrote: taking modulo 8 gives that $a,b,c$ are all odd. let $a \leq b \leq c$ so $3^a(3^{c-a}+3^{b-a}+1 )=x^2$ but $v_3(x^2)=a$ which is odd unless $a=b=c$ which is a solution. Wait i don't understand Taking modulo 8 gives $(-5)^a + (-5)^b + (-5)^c \equiv 3^a + 3^b + 3^c = x^2 \pmod {8}$ If $a,b,c$ all odd, hence it's negative, but $x^2$ never negative.

Jason99 wrote: Yaghi wrote: taking modulo 8 gives that $a,b,c$ are all odd. let $a \leq b \leq c$ so $3^a(3^{c-a}+3^{b-a}+1 )=x^2$ but $v_3(x^2)=a$ which is odd unless $a=b=c$ which is a solution. Wait i don't understand Taking modulo 8 gives $(-5)^a + (-5)^b + (-5)^c \equiv 3^a + 3^b + 3^c = x^2 \pmod {8}$ If $a,b,c$ all odd, hence it's negative, but $x^2$ never negative. $x^2 \equiv {0,1,4} \pmod {8}$ $3^k \equiv {1,3} \pmod {8}$ $3^a+3^b+3^c=y^2 \implies$ $a,b,c$ are all odd.

A similar question from Vietnam Olympiad 1982: Find all positive integral triplets (p, q, r) such that 2^p + 2^q + 2^r = 2336.

Although 2336 is not a perfect square but the approach is similar

you all are wrong the correct answer is a=b=c=2k+1

Yes ı agree aksshatkhanna

3|x^2 then 3|x x=3k and from some equations answer cames (sorry ı don't know how the LaTeX uses )

Let $a$ be the minimum vale over all positive integers such that $3^a+3^b+3^c=k^2$ for some positive integer $k$. Obviously, $c=1$ or $2$. Case $1$: Suppose there exists a value $k$ for $c=1.$ $3^a+3^b+3^c=k^2 \implies 3^a+3^b+3=k^2 \implies 3|k$ so $3^{a-1}+3^{b-1}+1= 3m^2$ for some positive integers $a-1, b-1,$ and $m$. This also means $$3^{a-1}+3^{b-1}=3m^2-1.$$Since $3$ does not divide $3m^2-1$ for positive integer $m$, $3^{b-1}=1 \implies b=1.$ It follows that $3^{a-1}=3m^2-2$ and $a=1$. $(1,1,1)$ is a solution to the equation. Case $2$: Suppose there exists a value $a$ where $c=2.$ Then $3^a+3^b+3^c=k^2 \implies 3^{a-2}+3^{b-2}+1=m^2$ for some positive integer $m$, and $3^{a-2}+3^{b-2}=m^2-1$. Then $3^{a-2}+3^{b-2}=(m+1)(m-1)$. We now separate this case into $2$ cases. Case $2a$: Let $3^k=m-1$ for some $k \leq b-2,$ and $$3^{k+2}=3^{a-k+2}+3^{b-k+2} $$$(1)$. When writing the base $3$ representation of (1), we see that this is impossible for an integer $k.$ Case $2b$: Let $3^k=m+1$ while $3^k-2=m-1=3^{a-k-2}+3^{b-k-2}$ for $k \leq b-2.$ This implies that $2=3^k-3^{a-k-2}-3^{b-k-2} \implies 1= (3^k -1)-3^{a-k-2}-3^{b-k-2}$. In base 3, this is $$1_3= 22222 \cdots 2_3-1 \cdots 0_3- 1 \cdots 1_3$$This is clearly impossible as well. Therefore, the only base case is when $a=1$ and $(a,b,c)=(1,1,1).$ In other words, $3^1+3^1+3^1=9=k^2$ is a base case solution. By multiplying each side by $3^{2k}$ for some positive integer $k$, we get that all $(a,b,c)=(2k+1, 2k+1, 2k+1)$ satisfy the problem statement.

If a=b=c then, 3^a+3^a+3^a= 3.3^a= 3^(a+1) If a=2k-1 then 3^(2k+1-1)= 3^(2k)= (3^k)^2 it is complete square so there are infinite triplets.

$\color{blue} \boxed{\textbf{SOLUTION}}$ $x^2 \equiv {0,1,4} \pmod {8}$ $3^k \equiv {1,3} \pmod {8}$ $3^a+3^b+3^c=y^2 \implies a,b,c$ are all odd. $3^a(3^{c-a}+3^{b-a}+1 )=x^2$ WLOG, $a \leq b \leq c$ $\textbf{Case 1:}$ If atleast two of $a,b,c$ are different WLOG, $a,b$ are different and $ a < b$ Then, $3|3^{b-a} \implies 3^{c-a}+3^{b-a}+1 \equiv 1,2 \pmod 3$ So, maximum power of 3 that divides $3^a(3^{c-a}+3^{b-a}+1 )$ is $a$ $\implies 3^a|x^2, 3^{a+1}$ doesn't divide $x^2$ but $a$ is odd which isn't possible [primes pairwisely stay in a square number that is in an even number of times] If all three are different then, $3^{c-a}+3^{b-a}+1 \equiv 1 \pmod 3$ And again $a$ is odd $\implies$ no $x$ $\textbf{Case 2:}$ If $a=b=c=$odd then $3^{a}+3^{b}+3^{c}=3^{a+1}=[3^{\frac{a+1}{2}}]^{2}$ So, $\color{red} \boxed{\textbf{(a,b,c)=(2t+1,2t+1,2t+1)}}$