Ta có:
\(\hept{\begin{cases}\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}\\\left(\sqrt{2}-\sqrt{3}\right)^2=5-2\sqrt{6}\end{cases}}\)
Ta chứng minh: Với mọi \(n\in N;n>0\)thì \(\left(5+2\sqrt{6}\right)^n+\left(5-2\sqrt{6}\right)^n\in Z\)
Với \(n=1\)thì \(\left(5+2\sqrt{6}\right)^1+\left(5-2\sqrt{6}\right)^1=10\in Z\)
Với \(n=2\)thì \(\left(5+2\sqrt{6}\right)^2+\left(5-2\sqrt{6}\right)^2=98\in Z\)
Giả sử nó đúng đến \(n=k\)hay
\(\left(5+2\sqrt{6}\right)^k+\left(5-2\sqrt{6}\right)^k=a\in Z\)
Ta chứng minh nó đúng với \(n=k+1\) hay \(\hept{\begin{cases}\left(5+2\sqrt{6}\right)^{k-1}+\left(5-2\sqrt{6}\right)^{k-1}=a\in Z\\\left(5+2\sqrt{6}\right)^k+\left(5-2\sqrt{6}\right)^k=b\in Z\end{cases}}\)
Ta có:
\(\left(5+2\sqrt{6}\right)^{k+1}+\left(5-2\sqrt{6}\right)^{k+1}\) \(=\left(5+2\sqrt{6}\right).\left(5+2\sqrt{6}\right)^k+\left(5-2\sqrt{6}\right).\left(5-2\sqrt{6}\right)^k\)
\(=\left(5+2\sqrt{6}\right).\left(b-\left(5-2\sqrt{6}\right)^k\right)+\left(5-2\sqrt{6}\right).\left(b-\left(5+2\sqrt{6}\right)^k\right)\)
\(=b\left(\left(5+2\sqrt{6}\right)+\left(5-2\sqrt{6}\right)\right)-\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)^k-\left(5-2\sqrt{6}\right).\left(5+2\sqrt{6}\right)^k\)
\(=10b-\left(5-2\sqrt{6}\right)^{k-1}-\left(5+2\sqrt{6}\right)^{k-1}\)
\(=10b-a\in Z\)
Vậy theo quy nạp thì nó đúng.
Quay lại bài toán thì ta có:
\(\left(\sqrt{2}+\sqrt{3}\right)^{2310}+\left(\sqrt{2}-\sqrt{3}\right)^{2310}=\left(5+2\sqrt{6}\right)^{1155}+\left(5-2\sqrt{6}\right)^{1155}\in Z\)