1977 Canadian MO Problems/Problem 1: Difference between revisions
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Directly plugging <math>\displaystyle a</math> and <math>\displaystyle b</math> into the function, <math>\displaystyle 4a^2+4a=b^2+b.</math> We now have a quadratic in <math>\displaystyle a.</math> | Directly plugging <math>\displaystyle a</math> and <math>\displaystyle b</math> into the function, <math>\displaystyle 4a^2+4a=b^2+b.</math> We now have a quadratic in <math>\displaystyle a.</math> | ||
Applying the quadratic formula, | Applying the quadratic formula, <math> \displaystyle a=\frac{-1\pm \sqrt{b^2+b+1}}{2}. </math> | ||
Because, <math>\displaystyle b^2< b^2+b+1 <(b+1)^2,</math> the quantity <math>\displaystyle b^2+b+1</math> cannot be a perfect square when <math>\displaystyle b</math> is an integer. Hence, when <math>\displaystyle b</math> is a positive integer, <math>\displaystyle a</math> cannot be. | Because, <math>\displaystyle b^2< b^2+b+1 <(b+1)^2,</math> the quantity <math>\displaystyle b^2+b+1</math> cannot be a perfect square when <math>\displaystyle b</math> is an integer. Hence, when <math>\displaystyle b</math> is a positive integer, <math>\displaystyle a</math> cannot be. | ||
== See also == | == See also == | ||
Revision as of 01:01, 22 July 2006
Problem
If 
prove that the equation 
has no solutions in positive integers 
and 
Solution
Directly plugging and 
into the function, 
We now have a quadratic in 
Applying the quadratic formula, 
Because, 
the quantity 
cannot be a perfect square when is an integer. Hence, when is a positive integer, cannot be.
