6. How many natural numbers $n$ exist for which $4^{n}-15$ is a square of an integer?
Answer. Two.

Solution. Let $4^{n}-15=x^{2}$, where $x$ is an integer. It is obvious that $x \neq 0$. If $x$ is negative, then $(-x)^{2}$ is also equal to $4^{n}-15$; therefore, we will assume that $4^{n}-15=x^{2}$, where $x$ is a natural number. From the equation $2^{2 n}-15=x^{2}$, we get: $2^{2 n}-x^{2}=15$, and using the formula for factoring the difference of squares: $\left(2^{n}-x\right)\left(2^{n}+x\right)=15$. Since $x$ is a natural number, the second factor on the left in the last equation is positive, and therefore the first factor must also be positive. The number 15 can be factored into natural numbers in two ways: $15=3 \cdot 5=1 \cdot 15$. Since $x>0$, we have $2^{n}+x>2^{n}-x$. Thus, there are only two possible cases:

$\left\{\begin{array}{l}2^{n}-x=1, \\ 2^{n}+x=15\end{array}\right.$ or $\left\{\begin{array}{l}2^{n}-x=3, \\ 2^{n}+x=5 .\end{array}\right.$

Solving the first system of equations (it is most convenient to simply add the equations), we get that $2 \cdot 2^{n}=16$, i.e., $n=3$. Similarly, from the second system, we get that $n=2$.

We could solve without limiting ourselves to natural values of $x$, but then the number of systems to be considered would increase, as the cases $15=(-1) \cdot(-15)=(-3) \cdot(-5)$ would also be possible.

Grading criteria.

- Correct answer with full justification - 7 points.
- Cases of factoring 15 into negative factors not considered without justification, why negative factors can be ignored - 6 points.
- Systems for determining $n$ correctly set up, but not checked for natural solutions - 5 points.
- Only one of the two possible factorizations of the number 15 considered - 5 points.
- Reasonable arguments that did not lead to a solution - 1-2 points.