# solution to $7^{a}+1 =3^{b}+5^{c}$ for natural $a$,$b$ and $c$

How do I solve $7^{a}+1 =3^{b}+5^{c}$ for natural $a$,$b$ and $c$?All I got after some modular arithmetic is that the $a$,$b$ and $c$ are all odd.The problem was posted on Art of Problem Solving(with no responses till now) and is supposedly from India International Mathematical Olympiad Training Camp.I was wondering if someone could please shed some insight as to whether it can be solved.

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Equations like this were discussed in a series of papers by Leo Alex and Lorraine Foster some years ago. You might have a look for them. One that looks relevant is Leo J. Alex, The Diophantine equation $3^a+5^b=7^c+11^d$, Math. Student 48 (1980), no. 2-4, 134–138 (1984), MR0776728 (86e:11022). –  Gerry Myerson Apr 5 '12 at 5:21
Can you please provide a link? –  Eisen Apr 5 '12 at 5:31
I don't have a link to provide. Suggest you see what you can find via Google and/or university library if you have access to one. –  Gerry Myerson Apr 5 '12 at 5:59
Don't know, whether this leads to somewhere... but perhaps it is an improvement to think of a=b=c=1 as a first solution and then to consider $\small 7^a-7^1 +1-1 = 3^b-3^1 + 5^c-5^1$ and then $\small 7(7^A -1) = 3(3^B-1) + 5(5^C-1)$ where A=a-1,B=b-1,C=c-1. Sometimes such an approach has helped me solving related problems. –  Gottfried Helms Apr 5 '12 at 8:00
@Raymond, Mathematics is eternal. –  Gerry Myerson Sep 2 '12 at 13:04

It looks as if the problem can be conquered by persistent use of modular arithmetic. Throughout, it is assumed that $a,b,c$ are sufficiently large.

1. Apply $\bmod{3}$ to find that $$2 \equiv 2^c \pmod{3}$$ It follows that $c \equiv 1 \pmod{2}$. Write $c = 2c_1 + 1$.
2. Apply $\bmod{8}$ to find that: $$(-1)^a + 1 \equiv 3^b + 5 \pmod{8}$$ It follows that $a \equiv b \equiv 1 \pmod{2}$. Write $a = 2a_1 + 1, \ b = 2b_1 + 1$.
3. Apply $\bmod{5}$ to find that: $$2 \cdot (-1)^{a_1} +1 \equiv 3 \cdot (-1)^{b_1} \pmod{5}$$ It follows that $a_1 \equiv b_1 \equiv 0 \pmod{2}$. Write $a_1 = 2a_2,\ b_1 = 2b_2$. Th
4. Apply $\bmod{16}$ to find that: $$7+1 \equiv 3 + 5 \cdot 9^{c_1} \pmod{16}$$ It follows that $c \equiv 0 \pmod{2}$. Write $c_1 = 2c_2$. The equation is now: $$7^{4a_2 + 1} + 1 = 3^{4b_2+1} + 5^{4c_2+1}$$

5. Apply $\bmod{13}$ [note: the number $13$ comes from the observation that $5^4 \equiv 1 \pmod{13}$ and $\phi(13) = 12$ is divisible by $4$] to find that: $$7^{4a_2 + 1} + 1 \equiv 3^{4b_2+1} + 5 \pmod{13}$$ By Fermat's theorem, only the residues of $4a_2 + 1$ and $4b_2+1$ in arithmetic $\bmod \phi(13)$ play a role in the above congruence, so we have just $\frac{\phi(13)}{\gcd(4,\phi(13))} = 3$ values of $a_2,b_2$ to check. It turns out that $a_2,b_2 \equiv 0 \pmod{3}$ are the only solutions. Write $a_2 = 3 a_3$ and $b_2 = 3b_3$. The equation is now: $$7^{12a_3 + 1} + 1 = 3^{12b_3+1} + 5^{4c_2+1}$$

6. Apply $\bmod{9}$ to find that: $$7 + 1 \equiv 5\cdot4^{c_2} \pmod{9}$$ By checking the $6$ possible values of $c_2$, one finds that $c_2 \equiv 2 \pmod{3}$. Write $c_2 = 3c_3 + 9$. The equation is now: $$7^{12a_3 + 1} + 1 = 3^{12b_3+1} + 5^{12c_3+9}$$

7. Apply $\bmod{37}$. Again, it is a reasonable guess because $\phi(37) = 36$ has many common factors with the $12$ that occurs in the exponentials, so the resulting congruence depends only on $a_3,b_3,c_3 \pmod{3}$. A rather mudane [I advise against attempting to do it by hand] direct check proves that there are no solutions.

8. We have shown that there are no solutions with the assumption that all numbers are "sufficiently large". One should now check how large the numbers really have to be, and look for small solutions. This will be significantly easier, since at least one of $a,b,c$ will be fixed at some small value. However, a lot of cases will be involved.

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