Learn / Library / Mental math

Representations as a gap of squares

Mental mathHard

In how many ways can $45$ be written as $a^{2} - b^{2}$ , where $a$ and $b$ are non-negative integers with $a > b$ ? Different pairs $(a, b)$ count as different ways.

Checking an answer needs JavaScript. The hint and the solution open without it.

Hint

Factor $45$ as $(a - b) (a + b)$ and think about parity.

Solution

Since $a^{2} - b^{2} = (a - b) (a + b) = 45$ , we need factor pairs $p \times q = 45$ with $p < q$ .
The pairs are $1 \times 45$ , $3 \times 15$ and $5 \times 9$ , all with both factors odd, so $a = \frac{p + q}{2}$ and $b = \frac{q - p}{2}$ are integers.
These give $(23, 22)$ , $(9, 6)$ and $(7, 2)$ .
So there are $3$ ways.

Answer: $3$

Next question Back to the library