Difference between revisions of "1971 Canadian MO Problems/Problem 6"
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=== Solution 2 === | === Solution 2 === | ||
+ | |||
+ | Assume that <math>n^2+2n+12=121k</math> for some integer <math>k</math> then | ||
+ | <cmath>n^2+2n+(12-121k)=0</cmath> | ||
+ | <cmath>\begin{align*} | ||
+ | x&=\frac{-2\pm\sqrt{4-4(12-121k)}}{2} \ | ||
+ | &=\frac{-2\pm2\sqrt{484k-44}}{2} \ | ||
+ | &=\sqrt{11(11k-1)} \ | ||
+ | \end{align*}</cmath> | ||
+ | By the assumption that <math>n</math> is an integer, <math>11k-1</math> must has a factor of <math>11</math>, which is impossible, contradiction. | ||
+ | |||
+ | ~ Nafer | ||
== See Also == | == See Also == | ||
{{Old CanadaMO box|num-b=5|num-a=7|year=1971}} | {{Old CanadaMO box|num-b=5|num-a=7|year=1971}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Revision as of 19:18, 10 January 2020
Contents
[hide]Problem
Show that, for all integers ,
is not a multiple of
.
Solutions
Solution
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 |