1971 Canadian MO Problems/Problem 6
Show that, for all integers , is not a multiple of .
Notice . For this expression to be equal to a multiple of 121, would have to equal a number in the form . Now we have the equation . Subtracting from both sides and then factoring out on the right hand side results in . Now we can say and . Solving the first equation results in . Plugging in in the second equation and solving for , . Since * is clearly not a multiple of 121, can never be a multiple of 121.
Assume that for some integer then By the assumption that is an integer, must has a factor of , which is impossible, contradiction.
|1971 Canadian MO (Problems)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •||Followed by|