Difference between revisions of "1971 Canadian MO Problems/Problem 6"
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− | If <math>n \equiv 10 \mod 11</math>, <math>(n+1)^2 \equiv (10+1)^2 \equiv | + | If <math>n \equiv 10 \mod 11</math>, <math>(n+1)^2 \equiv (10+1)^2 \equiv 0\mod 11</math>. However, considering the equation <math>\mod 121</math> for <math>n \equiv 10 \mod 11</math>, testing <math>n = 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 120</math>, we see that <math>(n+1)^2 + 11 always leave a remainder of greater than 1 \mod 121</math>. |
Thus, for any integer <math>n</math>, <math>n^2+2n+12</math> is not a multiple of <math>121</math>. | Thus, for any integer <math>n</math>, <math>n^2+2n+12</math> is not a multiple of <math>121</math>. |
Revision as of 22:12, 14 December 2011
Problem
Show that, for all integers , is not a multiple of .
Solution
. Consider this equation mod 11. . The quadratic residues are , and (as shown below).
If , , thus not a multiple of 11, nor 121.
If , , thus not a multiple of 11, nor 121.
If , , thus not a multiple of 11, nor 121.
If , , thus not a multiple of 11, nor 121.
If , , thus not a multiple of 11, nor 121.
If , , thus not a multiple of 11, nor 121.
If , , thus not a multiple of 11, nor 121.
If , , thus not a multiple of 11, nor 121.
If , , thus not a multiple of 11, nor 121.
If , , thus not a multiple of 11, nor 121.
If , . However, considering the equation for , testing , we see that .
Thus, for any integer , is not a multiple of .
1971 Canadian MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |