## Solution The problem of this contest was to proof an equivalence: ![](https://steemitimages.com/p/NTy4GV6ooFRkjTArCrebYc2WCCmX2KY4SfTbUDpHWg6ZYGGDQxkSJMEeDstbGRGui2ebEURsduenEkUUxQ1WAp1yvfxKEXKyyX4zMKofsf5yNW7Pd9n6zTpc6CpRwT9kgr9wTfq9tgpfHYz7DKGgQ1sUYSTA6e52k49nouuG?format=match&mode=fit&width=640) I also just noticed that this inequation only holds for uneven, positive values of n(which is probably the reason why I had no entries this week). First of all a small explanation why this is even true: if x is bigger than 0: Then 1/x^n is only < 1 if x > 1. if x is smaller 0: Then 1/x^n is only < -1 if x > -1 These two cases connect both inequations. In this case you can rearrange the leftern inequation to: `1 < x^(n+1) if x > 0` → `x > 1` `1 > x^(n+1) if x < 0` → `1 > ±x` → `-1 < ±x` → `-1 < x` The ± comes from taking an even root. So if x > 0 than x > 1 is the only solution and if x < 0, x > -1 is the only solution. Putting the 2 together gets the rightern side of the equivalence: `x > 1` or `-1 < x < 0` And the way back to prove it actually is an equivalence: `x > 1` → `x^(n+1) > 1` → `x > 1/x^n` `-1 < x < 0` → `1 < x^(n+1)` → `x > 1/x^n` *↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓* ### List of participants with their entries: Name | solution found | comment - | - | - @tonimontana | none | found one of the errors I made → you get the reward *↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓* ## Winner draw: Not necessary since only 1 participant: Congratulations @tonimontana , you won 2 SBI! *↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓* ### The next contest starts soon. Don't miss it! ### This contest will continue as a 2-week contest as it's getting harder to find a unique problem and I have more stress lately.

quantumdeveloper