3. Find all natural numbers $n$ for which the number $2^{10}+2^{13}+2^{14}+3 \cdot 2^{n}$ is a square of a natural number.

$(16$ points)
# Solution.

8) Let $n>12$. Then $N=2^{10}\left(1+2^{3}+2^{4}+3 \cdot 2^{n-10}\right)=2^{10} \cdot(2 k+1)^{2}$, and $25+3 \cdot 2^{n-10}=4 k^{2}+4 k+1$, $3 \cdot 2^{n-10}=4 k^{2}+4 k-24, \quad 3 \cdot 2^{n-12}=k^{2}+k-3 \cdot 2^{n-12}=(k+3)(k-2)$, The numbers $k+3$ and $k-2$ have different parities, so one of them is a divisor of 3. Since $k>0$, either $k-2=1$, $k=3,3 \cdot 2^{n-12}=6, \quad n=13$, or $k-2=3, k=5,3 \cdot 2^{n-12}=24, \quad n=15$.

Answer: $13,15$.