P and Q are two positive integers such that P² = Q² +13. The product of the numbers P and Q is

The correct option is 42.

Let's find the values of the positive integers $P$ and $Q$ that satisfy the given equation:

$P^2=Q^2+13$

We can rearrange this equation by subtracting $Q^2$ from both sides to group the squared terms together:

$P^2-Q^2=13$

The left side of the equation is a difference of squares, which can be factored using the identity $a^2-b^2=(a-b)(a+b)$:

$(P-Q)(P+Q)=13$

Since $P$ and $Q$ are positive integers, both $(P-Q)$ and $(P+Q)$ must also be integers. Furthermore, because $P$ and $Q$ are positive, we know that:

$P+Q>P-Q$

The number 13 is a prime number, so its only positive integer factors are 1 and 13. Therefore, the product of the two integer factors $(P-Q)$ and $(P+Q)$ can only be written as:
$(P-Q)\times (P+Q)=1\times 13$

Matching the smaller factor to 1 and the larger factor to 13 gives us the following system of linear equations:

$P-Q=1$

$P+Q=13$

To solve for $P$, we add the two equations together:

$(P-Q)+(P+Q)=1+13$
$2P=14$
$P=7$

To find $Q$, we substitute $P=7$ back into the second equation:

$7+Q=13$
$Q=6$

Now we calculate the product of the two numbers $P$ and $Q$:

$P\times Q=7\times 6=42$

Thus, the product of the positive integers $P$ and $Q$ is indeed 42.