Let [MATH]\sqrt{n+\sqrt{1996}}-\sqrt{n-1}=k[/MATH], where [MATH]k[/MATH] is a positive integer.
[MATH]\sqrt{n+\sqrt{1996}}=k+\sqrt{n-1}[/MATH]
Showing posts with label checking it for each case. Show all posts
Showing posts with label checking it for each case. Show all posts
Find all positive integers [MATH]n[/MATH] for which [MATH]\sqrt{n+\sqrt{1996}}[/MATH] exceeds [MATH]\sqrt{n-1}[/MATH] by an integer. (First Solution)
Find all positive integers [MATH]n[/MATH] for which [MATH]\sqrt{n+\sqrt{1996}}[/MATH] exceeds [MATH]\sqrt{n-1}[/MATH] by an integer.
My solution:
Let [MATH]\sqrt{n+\sqrt{1996}}-\sqrt{n-1}=k[/MATH], where [MATH]k[/MATH] is a positive integer.
[MATH]\sqrt{n+\sqrt{1996}}=k+\sqrt{n-1}[/MATH]
Let [MATH]\sqrt{n+\sqrt{1996}}-\sqrt{n-1}=k[/MATH], where [MATH]k[/MATH] is a positive integer.
[MATH]\sqrt{n+\sqrt{1996}}=k+\sqrt{n-1}[/MATH]
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