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# No triples? :(

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

Note by Paramjit Singh
2 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 · 2 years, 8 months ago

Nice. · 2 years, 8 months ago

Comment deleted Feb 15, 2014

$$2^3 - 2^2 = 4 \equiv 1 \pmod {3}$$. You need to rethink your solution · 2 years, 8 months ago