kunny wrote: Find all ordered triplets of positive integers $(a,\ b,\ c)$ such that $2^a+3^b+1=6^c$. We see that $2^a \equiv 2 \pmod{3}$ so $a$ is odd. Case 1. If $a=1$ then $3^b+3=6^c$ implies $b=c=1$. Case 2. If $a \ge 3$ then $c \ge 2$, we obtain $4|6^c$ and $8|2^a$. It follows that $3^b \equiv 3 \pmod{4}$. We get $b$ is odd. Thus, $3^b \equiv 3 \pmod{8}$. Hence $2^a+3^b+1 \equiv 4 \pmod{8}$. From here we obtain $c=2$, so $2^a+3^b=35$. We find $a=3,b=3$ and $a=5,b=1$. Thus, $\boxed{(a,b,c)=(1;1;1),(3;3,2),(5;1;2)}$.

hey tangents be careful there should be a=5 . in third answer...!

Refer to here for a longer more detailed analysis into my thinking on this problem It's basically Tangent's solution in a bit more detail plus a lot of my work and thinking I put into the problem.

It is easy to observe that $a \ge 3,c \ge 3 \implies 8|2^a-6^c \implies 8|3^b+1$ which is impossible since all possible residues of $3^b$ modulo $8$ are $1,3$. For $a \ge 3,c=2$ we have $2^a+3^b=35 \implies (a,b)=(3,3),(5,1)$ For $a \ge 3,c=1$ there's no solution. For $a=2$ there's no solution and for $a=1$ the only is $(1,1,1)$ Thus $(a,b,c)=(1,1,1),(3,3,2),(5,1,2)$.

assume cis greater than or equal to 4 and take mod 16 by taking the inferences got(like a is odd etc) you geta contradiction hence job to check the small no.

Easy,easy

Sketch of the solution Taking both sides of equation $\mod 8$, we get $c\le 3$. Checking cases $c=1, c=2$, we get solutions $(a, b, c)=(1, 1, 1), (3, 3, 2), (5, 1, 2)$

Kunihiko_Chikaya wrote: Find all ordered triplets of positive integers $(a,\ b,\ c)$ such that $2^a+3^b+1=6^c$. Think this prob is too easy for Japan P2. If we say $a\ge3, c\ge3$ we get $2^a+3^b+1\equiv3^b+1\equiv0(mod8)$ For all $3^b$, $3^b\equiv3, 1(mod8)$ . Therefore, at least 1 of $a, c$ has to be smaller than 3. $1. a=2$ Then we get $3^b+5=6^c$ but this is not true because $3$ is not a divisor of $3^b+5$. $2. a=1$ Then we get$3^b+3=6^c$ and if we see $v_3(n)$ at each sides we can get only 1 solution $(a, b, c)=(1, 1, 1)$. $3. c=2 or c=1$ Then we get $2$ equations $2^a+3^b+1=6, 36$ and if we solve this we get $(a, b, c, d)=(3, 3, 2), (5, 1, 2)$ Therefore, $\boxed{(a,b,c)=(1,1,1),(3,3,2),(5,1,2)}$

Fun fact: this problem appeared in the manga "Mathematics Golden" (also known as Suugaku Golden), along with many other actual olympiad problems. I never read it myself (only saw snippets, including this problem), but from what I've seen the manga has a surprisingly relatable depiction of both the problem-solving process and the community as a whole, including someone solving a problem during a bathroom break.

Okay, we go $\pmod{8}$, so note that if $a>2$ and $c>2$ then we have $3^b \equiv 7\pmod{8}$, note this is impossible so if $a=1$, then $b=c=1$ works, if not, we have $c=2$, and $2^a+3^b=35$, so $\boxed{a,b,c=(1,1,1) \text{or} (3,3,2) \text {or} (5,1,2)}$

Notice that if we examine $\pmod 3$ we obtain $2^a+3^b+1\equiv(-1)^a+1\pmod 3$ which forces $a$ to be odd. Case 1: $a=1$ Plugging $a=1$ we obtain $3+3^b=6^c\Longrightarrow1+3^{b-1}=3^{c-1}\cdot2^c$ however notice that $1+3^{b-1}\equiv1\pmod 3$ which forces $3^{c-1}\cdot2^c\equiv1\pmod 3$ which in turn implies $c=1$ which forces $b=1$ So one solution is $(a,b,c)=(1,1,1)$ So from now on assume $a\ge3\text{ and }c\ge2$ Notice that when we observe $\pmod 4$ we obtain $2^a+3^b+1\equiv(-1)^b+1\pmod 4$ which forces $b$ to be odd since $6^c\equiv0\pmod 4$ Case 2: $a\ge3\text{ and }c\ge3$ Inspecting $\pmod 8$ we obtain $2^a+3^b\equiv2\text{ or }4$ which is clearly a contradiction as $6^c\equiv0\pmod 8$ for $c\ge3$, thus $c\le2$ however since we know that $c\ge2$, we must have $c=2$ Plugging this into our original equation we obtain $2^a+3^b=35$, furthermore since $a\ge3$ we must have that $2^a\ge8$ which implies that $35\ge8+3^b\Longleftrightarrow27\ge3^b$ which forces either $b=1\text{ or }b=3$. Checking both of these cases we obtain $(a,b,c)=(3,3,2)\text{ and }(5,1,2)$ To sum up $\boxed{(a,b,c)=(1,1,1),(3,3,2)\text{ and }(5,1,2)\text{ are the only solutions for }a,b,c\in\mathbb{Z}^+}$ $\blacksquare$.

Case 1: $c=1\implies 2^a+3^b=5\implies a=b=1$ Case 2: $c=2\implies 2^a+3^a=35$ Clearly $b<4$ and checking $b=1,2,3$ we get that $a=3,b=3$ and $a=5,b=1$ are solutions Case 3: $c\geq 3\implies 6^c\equiv 0 \pmod 8$ $3^b\equiv 1,3\pmod 8$ Combining the two we get that $2^a+3^b+1\equiv 6^c\pmod 8\implies b$ is odd and $2^a\equiv 4\pmod 8\iff a=2$ $\implies 6^c-3^b=5$ which is a contradiction $\pmod 3$ $(a,b,c)=(1,1,1);(3,3,2);(5,1,2)$