Difference between revisions of "1971 Canadian MO Problems/Problem 6"
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reveals that only <math>10</math> satisfy the condition <math>{n^{2} + 2n + 12} \equiv 0 \pmod{11}</math>. | reveals that only <math>10</math> satisfy the condition <math>{n^{2} + 2n + 12} \equiv 0 \pmod{11}</math>. | ||
− | Plugging <math>10</math> into <math>{n^{2} + 2n + 12}</math> shows that it is not divisible by 121. | + | Plugging <math>10</math> into <math>{n^{2} + 2n + 12}</math> shows that it is not divisible by <math>121</math>. |
+ | |||
+ | Thus, there are no integers <math>n</math> such that <math>n^{2} + 2n + 12</math> is divisible by <math>121</math>. | ||
~iamselfemployed | ~iamselfemployed |
Revision as of 16:00, 15 May 2024
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
Solution 3
In order for to divide
,
must also divide
.
Plugging in all numbers modulo :
or
to make computations easier,
reveals that only satisfy the condition
.
Plugging into
shows that it is not divisible by
.
Thus, there are no integers such that
is divisible by
.
~iamselfemployed
See Also
1971 Canadian MO (Problems) | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 7 |