Problem:
How many ordered pairs of integers satisfy the equation
Answer Choices:
A.
B.
C.
D.
E. infinitely many
Solution:
Rearranging the terms and completing the square for yields the result
Then, notice that can only be and because any value of that is greater than will cause the term to be less than , which is impossible as must be real. Therefore, plugging in the above values for gives the ordered pairs , and gives a total of ordered pairs.
Bringing all of the terms to the LHS, we see a quadratic equation
in terms of . Applying the quadratic formula, we get
In order for to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant,
must be non-negative. Therefore,
Here, we see that we must split the inequality into a compound, resulting in . The only integers that satisfy this are . Plugging these values back into the quadratic equation, we see that both produce a discriminant of , meaning that there is only solution for . If , then the discriminant is nonzero, therefore resulting in two solutions for .
Thus, the answer is .
The problems on this page are the property of the MAA's American Mathematics Competitions