Given that and are integers, find all possible solutions, i.e. pairs of such that the sum of the square of x and 615 is equal to two raised to the nth power or .

Source: mathcontest.olemiss.edu 10/31/2011

**SOLUTION
** implies that is odd which in turn implies that is odd. Let’s take a look at a few values of where is an odd integer:

The pattern shows that ends in 1, 5, or 9. Hence, it follows that ends in 6, 0, or 4 respectively.

Let’s take a look at a few values of where is an integer:

The non-positive integers produce too small a value compared to so we will ignore them. Instead, we investigate the positive integers which produce values that end in 2, 4, 8, or 6. For the equation to hold true, must end in either 4 or 6. It follows from the above pattern that must be even or for some positive integer .

We write

The possible products that produce 615 are:

Which one is the correct one? Let’s add the factors to figure out:

We now have a guideline to pick the correct product: the sum of its factors is a power of 2.

The sums of the factors are:

Not a power of 2

No

Yes a power of 2

No

We have

**Value of **

**Values of **

**Answer**: and .