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What are the positive integer solutions to the equation

$$2^a + 3^b = 5^c$$

Of course $(1,\space 1, \space 1)$ is a solution.

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Anything else would violate Beal's conjecture so the solutions given are likely it. –  EuYu Nov 1 '12 at 7:36
    
This result should solve the case that $a$ is even: "It is also true that the equation $3^x+4^y=5^z$ has no solutions in natural numbers except $x = y = z = 2$." Sierpinski, Elementary Number Theory, p.40. Refernces given there: W. Sierpinski: O rownaniu $3^x+4^y=5^z$, {Wiadom. Mat.}, 1 (1956), 194-195. Nagell T.: Sur une classe d'equations exponentielles, Ark. Mat. 3 (1958) 569-582. –  Martin Sleziak Nov 2 '12 at 18:59
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3 Answers 3

up vote 10 down vote accepted

If $a=0$, then it is clear there are no solutions. If $b=0$, then we need $2^a + 1 = 5^c$. It is easy to show in this case $a=2,c=1$ is the only solution by showing that we need $2^{a-2}|c$. When $c=0$ there are obviously no solutions.

Suppose $a=1$. Then $2 + 3^b = 5^c$ only has the solution of $b=1,c=1$. To show this check modulo $275$ to deduce $c=1$ and thus $b=1$ is forced.

Now, suppose $a \ge 3$. Then remark that by checking modulo $4$ we need $b$ to be even so let $b = 2b'$. So let's solve $2^a + 3^{2b'} = 5^c$. Checking modulo $8$ we get $c$ is even so let $c = 2c'$. Then: $$2^a = (5^{c'} - 3^{b'})(5^{c'}+3^{b'})$$ We get $5^{c'} - 3^{b'} = 2^m, 5^{c'}+3^{b'} = 2^n$ for some $m,n$. But then $2 \cdot 5^{c'} = 2^m + 2^n$, forcing $m=1$. Thus we need $5^{c'} - 3^{b'} = 2$. But we already showed this only has the solution $b' = 1, c' = 1$. Thus it follows the only solution where $a \ge 2$ is with $a=4, b = 2, c= 2$.

Putting every together, we have proven the only solutions are $(2,0,1), (1,1,1), (4,2,2)$

EDIT: I realize I forgot to do the case of $a=2$. So we need to solve $4 + 3^{2b'} = 5^c$. Modulo $275$ happens to work again to force $c=1$ and thus we get no solutions when $b$ is nonzero.

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What is the purpose of using modulo 275? Also, you should check out Størmer's theorem and Baker's theorem. They can simplify the above proof to maybe 2 paragraphs at most, and neither of them require modular arithmetic. –  Ryan Dec 13 '12 at 21:55
    
The purpose of modulo $275$ is to deduce $c=1$ (note that $25|275$). Also considering this problem is tagged as Elementary Number Theory, I don't think the question poser was looking for solutions using that kind of heavy machinery. –  dinoboy Dec 13 '12 at 23:06
    
Would it have worked if you used 175 instead? I'm still not understanding if 275 is an arbitrary choice or not. –  Ryan Dec 14 '12 at 1:10
    
275 is decently arbitrary, the reason why it works is because the order of 3 and 5 modulo 11 is fairly low since they are both quadratic residues so the chance of it working is much higher than other primes such as 7. –  dinoboy Dec 14 '12 at 1:46
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Another solution is $2^4 + 3^2 = 5^2$. That's probably all.

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Here is a little program in Python:

for k in range(10):
for l in range(10):
    for m in range(10):
        if 2**k + 3**l == 5**m:
            print (k, l, m)

Its output is as follows

    python check.py
1 1 1
2 0 1
4 2 2

This yields these three equations.

$$2 + 3 = 5$$ $$2^2 + 3^0 = 5^1$$ $$2^4 + 3^2 = 5^2.$$

And that's all folks! Swap 100 for 10 and the result is the same. Cooking up an analytical proof of we have done is not hard.

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the second solution contains a $0$ –  user31280 Nov 1 '12 at 0:23
    
C'est vrai. Ignore it. I just was checking the corner cases. –  ncmathsadist Nov 1 '12 at 0:54
    
I just checked that there are no other solutions for $a, b < 600$. –  Patrick Li Nov 1 '12 at 1:25
1  
Could there be solutions > 600? –  Caltor Nov 1 '12 at 13:53
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