Difference between revisions of "1971 Canadian MO Problems/Problem 6"
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Notice <math>n^{2} + 2n + 12 = (n+1)^{2} + 11</math>. For this expression to be equal to a multiple of 121, <math>(n+1)^{2} + 11</math> would have to equal a number in the form <math>121x</math>. Now we have the equation <math>(n+1)^{2} + 11 = 121x</math>. Subtracting <math>11</math> from both sides and then factoring out <math>11</math> on the right hand side results in <math>(n+1)^{2} = 11(11x - 1)</math>. Now we can say <math>(n+1) = 11</math> and <math>(n+1) = 11x - 1</math>. Solving the first equation results in <math>n=10</math>. Plugging in <math>n=10</math> in the second equation and solving for <math>x</math>, <math>x = 12/11</math>. Since <math>12/11</math> *<math>121</math> is clearly not a multiple of 121, <math>n^{2} + 2n + 12</math> can never be a multiple of 121. | Notice <math>n^{2} + 2n + 12 = (n+1)^{2} + 11</math>. For this expression to be equal to a multiple of 121, <math>(n+1)^{2} + 11</math> would have to equal a number in the form <math>121x</math>. Now we have the equation <math>(n+1)^{2} + 11 = 121x</math>. Subtracting <math>11</math> from both sides and then factoring out <math>11</math> on the right hand side results in <math>(n+1)^{2} = 11(11x - 1)</math>. Now we can say <math>(n+1) = 11</math> and <math>(n+1) = 11x - 1</math>. Solving the first equation results in <math>n=10</math>. Plugging in <math>n=10</math> in the second equation and solving for <math>x</math>, <math>x = 12/11</math>. Since <math>12/11</math> *<math>121</math> is clearly not a multiple of 121, <math>n^{2} + 2n + 12</math> can never be a multiple of 121. |
Revision as of 22:22, 18 January 2023
Contents
[hide]Problem
Show that, for all integers ,
is not a multiple of
.
Solutions
Solution 1
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.
Solution 2
Assume that for some integer
then
By the assumption that
is an integer,
must has a factor of
, which is impossible, contradiction.
~ Nafer
See Also
1971 Canadian MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |