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设\(n+10=ka^2\ \ \ 4n+7=kb^2\),消去\(n{,}\ \frac{kb^2-7}{ka^2-10}=4\),\(\therefore\ 33=1\times3\times11=k\left( 2a-b\right)\left( 2a+b\right)\),\(\therefore\ k=1{,}\ 2a-b=3{,}\ 2a+b=11;\)或者\(k=3{,}\ 2a-b=1{,}\ 2a+b=11\)或者\(k=11{,}\ 2a-b=1{,}\ 2a+b=3\)。这样可以求得\(k{,}\ a,b\),从而代入求n,所以最大可能值=17 |
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