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.
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