This equation does not have solutions modulo $ 80$. We have $ (10^a\bmod 80)\in\{ 10, 20, 40, 0 \}$ $ (2^b\bmod 80)\in\{ 2, 4, 8, 16, 32, 64, 48 \}$ $ (3^c + 1997\bmod 80)\in\{ 0, 6, 24, 78 \}$ but not two elements from the first two sets sum up to an element of the third. oops! See correction below.

how about $ 20+4=24$ ?

Happily, this still suffices as $ 10^a \equiv 20 \mod 80$ for exactly one $ a$ (namely $ 2$). Similar, $ 2^b \equiv 4 \mod 80$ has only one solution $ b=2$, thus this combination does not give a solution.

No, that's still wrong. $ 64 + 40 \equiv 24 \pmod{80}$.

Ok, so from the mod $ 80$ we see that we must have $ a=3$ and $ b \geq 6$. Thus, we are left with the equation $ 2^b - 997 = 3^c$. This equation DOES have the solution in case you haven't noticed that already, namely $ b=10$ and $ c=3$. It's also easy to check that for $ c=1$ and $ c=2$ there are no solutions, so let's assume that $ c \geq 4$. Order of $ 2$ mod $ 81$ is $ 54$ so the equation $ 2^x - 997 \equiv 2^x - 25 \equiv 0 \pmod{81}$ has the unique solution for $ x \pmod{54}$, namely $ x \geq 46 \pmod{54}$, thus $ b \equiv 46 \pmod{54}$, in particular we have $ b \geq 10 \pmod{18}$ and $ 2^b \equiv 2^{10} \equiv 17 \pmod{19}$. It follows that $ 3^c \equiv 8 \pmod{19}$ and since order of $ 3$ mod $ 19$ is $ 18$ this implies that $ c \equiv 3 \pmod{18}$. Now we do modulo $ 433 = 8 \cdot 27 + 1$ which is a prime number. It's easy to check that order of $ 2$ mod $ 433$ is $ 72$ and as we remember that $ b \equiv 10 \pmod{18}$ we have $ 2^b-997 \in \{2^{10} - 997, 2^{28}-997, 2^{46}-997\} = \{27, 6, 144\} \pmod{433}$. The order of $ 3$ mod $ 433$ is $ 27$ and $ 3^c \in \{27, 6, 144\} \pmod{433}$ implies that $ c \equiv 3 \pmod{27}$. Now, we do modulo $ 262657 = \frac{2^{27} - 1}{2^{9}-1}$. This number is prime (but unfortunately we have to verify this with computer). Since the order of $ 2$ is $ 27$ and we know that $ b \equiv 19 \mod{27}$ we have that $ 2^b - 997 \equiv 2^{19} - 997 \equiv 260634 \pmod{262657}$. And now luckily it happens that order of $ 3$ is $ 54$ and since we know that $ c \equiv 3 \pmod{27}$ we have only to check $ 3^3$ and $ 3^{30}$ mod $ 262657$. But this values are $ 27$ and $ 258554$ respectively, a contradiction. I guess that a contradiction could be reached also with some prime number in the interval $ [600, 2000]$. You can try to find some better number than $ 262657$ but I'm pretty sure that it can't be less than $ 600$.

Sorry for confusion. Modulo $ 80$ works but explanation is a bit more involved: $ 10^a\bmod 80: 10, 20, 40, [0]$ (in brackets is the period) $ 2^b\bmod 80: 2, 4, 8, [16, 32, 64, 48]$ $ 3^c + 1997\bmod 80: [0, 6, 24, 78]$ It is clear that there is no solution in the periods of these functions. While the remaining cases is easy to check manually (with the help of computers).

You are still wrong! Modulo $ 80$ gives us only that $ a=3$ and it's not easy to check this manually. In fact, the whole difficulty of this problem is the case $ a=3$.

Well, it may not be easy to test manually (i.e., without computer) but anyway there is an easy algorithm for such problems. See http://www.mathlinks.ro/viewtopic.php?t=48431 For $ a = 3$ the problem is to solve $ 2^b = 3^c + 997$. My software easily produces the number $ 42143976$ (minimality was not aimed) modulo which there is no periodic solution. TomciO wrote: I guess that a contradiction could be reached also with some prime number in the interval $ [600, 2000]$. Just for curiosity I've tested all moduli below $ 9000$ - no luck.

$ 2^b=3^c+997\mod p$ had not solution for $ p=73$ and for $ p=601$.

maxal your number is pretty good since it is only $ 8 \cdot 243 \cdot 19 \cdot 163$ so it doesn't contain any big prime factor. It is probably an advantage of doing modulo $ 243$ instead of $ 81$. Rust, this congruences have solution: $ b = 10, c = 3$. There is no possibility to obtain a contradiction without using that $ b \equiv 46 \pmod{54}$ or $ c \equiv 259 \pmod{512}$, i.e. we have to choose numbers with $ \varphi$ divisible by at least $ 27$ or a big power of two. So $ 73$ and $ 601$ aren't best choices.

Yes, $ b=10,c=3$ unique solution. Consider by mod $ 73$ we get $ b=10\mod 9,c=3\mod 12$. Consider by mod $ 271$ we get $ b=10\mod 135,c=3\mod 30$ or $ b=19\mod 135 , c=27\mod 30$. Consider by mod $ 30$ we get $ b=0\mod 5,c=3\mod 30$ or $ b=2\mod 5,c=15\mod 30$. Therefore $ c=3\mod 60$ and $ b=10\mod 135$. If $ b=10\mod 135$, then $ 81\not |2^b-997$. Therefore $ c\le 3$

Rust I think that you've meant considering mod $ 31$ instead of mod $ 30$. And you are right, mod $ 271$ works! I've thought that I had checked it but I probably have made mistake in calculations. Your solution can be also little simplified, because it is enough to consider mod $ 19$ instead of mod $ 73$.

TomciO wrote: Rust I think that you've meant considering mod $ 31$ instead of mod $ 30$. And you are right, mod $ 271$ works! I've thought that I had checked it but I probably have made mistake in calculations. Your solution can be also little simplified, because it is enough to consider mod $ 19$ instead of mod $ 73$. Yes $ mod 31$. I choose $ p$, suth that minimal period 2 $ 27|T_p(2)$. minimal of them $ 271$, suth that $ 3^3|T_p(2)=135.$