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Reduce each side mod 7 we have:
a^5 + 2 ≡ 0 (mod 7)
a^5 ≡ -2 ≡ 5 (mod 7)
What natural numbers have a quintic residue of 5 (mod 7)? Answer: only 3 (mod 7), so a ≡ 3 (mod 7)
So start by trying a=3. We get 3^5 = 243. Add 100, we get 343 which is equal to 7^3, so b=3.

RexxSchneider
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a^5 + 100 = 7^b => a^5 + 343 - 243 = 7^b
=> a^5 - 3^5 = 7^b - 7^3

shashankkatiha
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Sir, please help me to solving a problem, A circle with Center( Pai, e), Find the maximum and minimum number of Rational point on it.

AbdulHaiMi
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I learn a lot with you, thanks so much

hamidhammoudi
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It would be easier if we write 100 as 7^3 - 3^5

abhinandansharma
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a = 7k + 3 and a = 4k + 3 has infinite solutions of the form a = 28K + 3 where k = 0, 1, 2, 3...
How on Earth can u conclude K = 0?

sharathpr