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Author |
hbghlyj
Posted 2022-6-9 21:04
aops
Define the sequence $ ( s_{n})$ by $ s_{n} =\left(( 2n+1)^{2} -2\right)^{2} \ \ ( n\geq 0)$.
It is clear that $ ( s_{n})$ is strictly increasing.
Since $ s_{n} +s_{n+1} =2\left( 4n^{2} +8n+5\right)^{2}$, $ ( s_{n} +s_{n+1}) /2$ is a perfect square.
We can show that $ \gcd( s_{n} ,s_{n+1}) =1$ as follows :
It is sufficient to show that $ \gcd\left(( 2n+1)^{2} -2,( 2n+3)^{2} -2\right) =1$.
$ \gcd\left(( 2n+1)^{2} -2,( 2n+3)^{2} -( 2n+1)^{2}\right) =\gcd\left(( 2n+1)^{2} -2,8( n+1)\right)$
$ =\gcd\left(( 2( n+1) -1)^{2} -2,n+1\right) =\gcd\left( 1^{2} ,n+1\right) =1$.
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