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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.

Thanks.(I am a bit skeptical about this problem although I may be wrong)

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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).2012-04-05
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    Can you please provide a link?2012-04-05
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    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.2012-04-05
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    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.2012-04-05
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    Concerning @Gerry's fine answer see this paper ['Exponential Diophantine Equations'](http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.pjm/1102724775) and [related papers](http://projecteuclid.org/DPubS?verb=Display&version=1.0&service=UI&handle=euclid.pjm/1102724775&page=record).2012-09-02
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    (a little late I'll admit... ;-))2012-09-02
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    @Raymond, Mathematics is eternal.2012-09-02
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    @GerryMyerson: Yes of course !2012-09-02
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    Perhaps the best reference is L. L. Foster, Solution to problem E2749, Amer. Math. Monthly, 87 (2), (1980), 138.2017-07-25

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