Difference between revisions of "1971 Canadian MO Problems/Problem 6"
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<math>n^2 + 2n + 12 = (n+1)^2 + 11</math>. Consider this equation mod 11. <math> (n+1)^2 + 11 \equiv (n+1)^2 \mod 11</math>. | <math>n^2 + 2n + 12 = (n+1)^2 + 11</math>. Consider this equation mod 11. <math> (n+1)^2 + 11 \equiv (n+1)^2 \mod 11</math>. | ||
− | The quadratic residues mod 11 are 1, 3, 4, 5, 9, and 0 (as shown below). | + | The quadratic residues <math>mod 11</math> are <math>1, 3, 4, 5, 9</math>, and <math>0</math> (as shown below). |
If <math>n \equiv 0 \mod 11</math>, <math>(n+1)^2 \equiv (0+1)^2 \equiv 1\mod 11</math>, thus not a multiple of 11, nor 121. | If <math>n \equiv 0 \mod 11</math>, <math>(n+1)^2 \equiv (0+1)^2 \equiv 1\mod 11</math>, thus not a multiple of 11, nor 121. |
Revision as of 21:06, 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 ,
, thus a multiple of 11. However, considering the equation
,
, thus not a multiple of 121, even though it is a multiple of 11.
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 |