Let a and b be two odd positive integers. Prove that gcd(2a+1,2gcd(a,b)−1)=1.
I tried rewriting it to get gcd(22k+1+1,2gcd(2k+1,2n+1)−1), but I didn't see how this helps.
How to find limh→0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
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