Pizza Combination

With $n$ being the number of available toppings, the number of pizza varieties is

$\dbinom{n}{3} + \dbinom{n}{2} + \dbinom{n}{1} + \dbinom{n}{0} = \dfrac{n(n - 1)(n - 2)}{6} + \dfrac{n(n - 1)}{2} + n + 1 =$

$\dfrac{1}{6}(n(n - 1)(n - 2) + 3n(n - 1) + 6n + 6) = \dfrac{1}{6}(n(n - 1)(n + 1) + 6(n + 1)) =$

$\dfrac{1}{6}(n + 1)(n^{2} - n + 6)$ .

Next, for each layer of the lasagna there are $n + 1$ choices, (any one of the $n$ toppings plus the option of no topping), so since there are two layers the number of lasagna varieties is $(n + 1)^{2}$ . We thus require that

$\dfrac{1}{6}(n + 1)(n^2 - n + 6) - (n + 1)^{2} = 1265 \Longrightarrow (n + 1)(n^{2} - n + 6 - 6(n + 1)) = 7590 \Longrightarrow$

$(n + 1)(n^{2} - 7n) = 7590 \Longrightarrow n(n + 1)(n - 7) = 7590 = 2 \times 3 \times 5 \times 11 \times 23$ .

We now need to arrange the prime factors into a product of the form $n(n + 1)(n - 7)$ , which can be done if we set $n = 2 \times 11 = 22$ , in which case $n + 1 = 23$ and $n - 7 = 3 \times 5 = 15$ . As no other combination will work, we can conclude that there are $\boxed{22}$ available toppings.