My teacher said we can solve this using Euclid's division lemma but I have no clue how. I factored the numbers into $$2^{60}-1 = (2^{10}-1)(1+2^{10}+2^{20}+...+2^{50})$$ and $$2^{50}-1 = (2^{10}-1)(1+2^{10}+2^{20}+...+2^{40})$$ So I got the gcd as $2^{10}-1$. But I don't know if there can be some common factor in $(2^{10}-1)(1+2^10+2^{20}+...+2^{40})$ and $(2^{10}-1)(1+2^{10}+2^{20}+...+2^{50})$.
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2Hint : The gcd must also divide $2^{60}-2^{50}$, the difference of the numbers. – Peter Jun 16 '20 at 10:54
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https://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 – lab bhattacharjee Jun 16 '20 at 11:01