Difference between revisions of "1969 Canadian MO Problems/Problem 7"

m
(Useless links)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Show that there are no integers <math>\displaystyle a,b,c</math> for which <math>\displaystyle a^2+b^2-8c=6</math>.
+
 
 +
Show that there are no integers <math>a,b,c</math> for which <math>a^2+b^2-8c=6</math>.
  
 
== Solution ==
 
== Solution ==
Note that all [[perfect square]]s are equivalent to <math>\displaystyle 0,1,4\pmod8.</math>  Hence, we have <math>\displaystyle a^2+b^2\equiv 6\pmod8.</math>  It's impossible to obtain a sum of <math>\displaystyle 6</math> with two of <math>\displaystyle 0,1,4,</math> so our proof is complete.
 
  
----
+
Note that all [[perfect square]]s are equivalent to <math>0,1,4\pmod8.</math>  Hence, we have <math>a^2+b^2\equiv 6\pmod8.</math>  It's impossible to obtain a sum of <math>6</math> with two of <math>0,1,4,</math> so our proof is complete.
* [[1969 Canadian MO Problems/Problem 6|Previous Problem]]
+
 
* [[1969 Canadian MO Problems/Problem 8|Next Problem]]
+
== References ==
* [[1969 Canadian MO Problems|Back to Exam]]
+
 
 +
{{Old CanadaMO box|num-b=6|num-a=8|year=1969}}
 +
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 10:22, 29 April 2008

Problem

Show that there are no integers $a,b,c$ for which $a^2+b^2-8c=6$.

Solution

Note that all perfect squares are equivalent to $0,1,4\pmod8.$ Hence, we have $a^2+b^2\equiv 6\pmod8.$ It's impossible to obtain a sum of $6$ with two of $0,1,4,$ so our proof is complete.

References

1969 Canadian MO (Problems)
Preceded by
Problem 6
1 2 3 4 5 6 7 8 Followed by
Problem 8
Invalid username
Login to AoPS