Difference between revisions of "1971 Canadian MO Problems/Problem 6"
(→Solution) |
(→Solution 2) |
||
Line 8: | Line 8: | ||
=== 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]] |
Latest revision as of 20:18, 10 January 2020
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 |