Difference between revisions of "1977 Canadian MO Problems/Problem 1"
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If <math>\displaystyle f(x)=x^2+x,</math> prove that the equation <math>\displaystyle 4f(a)=f(b)</math> has no solutions in positive integers <math>\displaystyle a</math> and <math>\displaystyle b.</math> | If <math>\displaystyle f(x)=x^2+x,</math> prove that the equation <math>\displaystyle 4f(a)=f(b)</math> has no solutions in positive integers <math>\displaystyle a</math> and <math>\displaystyle b.</math> | ||
Revision as of 14:51, 24 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,
In order for both and
to be integers, the discriminant must be a perfect square. However, since
the quantity
cannot be a perfect square when
is an integer. Hence, when
is a positive integer,
cannot be.