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Prove that there are no triples \((a,b,c)\) of positive integers satisfying \(2^a - 5^b7^c = 1 \).

Note by Paramjit Singh 3 years, 8 months ago

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Suppose there exists a triple \( (a,b,c,) \)

Then \( 2^a \equiv 1 \pmod5 \)

Since \( ord(5) = 4\) we have \( a = 4a'\) for some \( a' \).

Rearranging, \(2^{4a'} - 1 = 5^b7^c\)

\( 2^{4a'} - 1 \equiv (2^2)^{2a'} - 1 \equiv 1^{2a'} - 1 \equiv 0 \pmod3 \)

But \(5^b7^c \not\equiv 0 \pmod3 \)

Therefore there exist no triples \( (a,b,c) \) which satisfy the given statement

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Nice.

Comment deleted Feb 15, 2014

\( 2^3 - 2^2 = 4 \equiv 1 \pmod {3} \). You need to rethink your solution

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TopNewestSuppose there exists a triple \( (a,b,c,) \)

Then \( 2^a \equiv 1 \pmod5 \)

Since \( ord(5) = 4\) we have \( a = 4a'\) for some \( a' \).

Rearranging, \(2^{4a'} - 1 = 5^b7^c\)

\( 2^{4a'} - 1 \equiv (2^2)^{2a'} - 1 \equiv 1^{2a'} - 1 \equiv 0 \pmod3 \)

But \(5^b7^c \not\equiv 0 \pmod3 \)

Therefore there exist no triples \( (a,b,c) \) which satisfy the given statement

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Nice.

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Comment deleted Feb 15, 2014

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\( 2^3 - 2^2 = 4 \equiv 1 \pmod {3} \). You need to rethink your solution

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