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

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== Problem ==
 
== Problem ==
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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>\displaystyle a,b,c</math> for which <math>\displaystyle a^2+b^2-8c=6</math>.
  
 
== Solution ==
 
== Solution ==
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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>\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.
  
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== References ==
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* [[1969 Canadian MO Problems/Problem 6|Previous Problem]]
 
* [[1969 Canadian MO Problems/Problem 6|Previous Problem]]
 
* [[1969 Canadian MO Problems/Problem 8|Next Problem]]
 
* [[1969 Canadian MO Problems/Problem 8|Next Problem]]
 
* [[1969 Canadian MO Problems|Back to Exam]]
 
* [[1969 Canadian MO Problems|Back to Exam]]
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[[Category:Intermediate Number Theory Problems]]

Revision as of 22:53, 5 September 2006

Problem

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

Solution

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

References

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