Question 600971
<pre>
{{{  sqrt ( (3n-5)/(n+1) )}}}

In order for that to equal to an integer,
what's under the radical must be a perfect square. 

We compare the fraction {{{(3n-5)/(n+1)}}} to the fraction

{{{3n/n}}}, and observe that {{{(3n-5)/(n+1)}}} has a smaller 
numerator and a larger denominator than {{{3n/n}} so it's less 
than {{{3n/n}}} which is 3. So {{{  (3n-5)/(n+1 )}}} is a perfect
square less than 3. 

The only perfect square less than 3 is 1, so what's under the 
radical must equal to 1.  So

{{{ (3n-5)/(n+1) }}} = 1

Multiply both sides by n+1

3n - 5 = n + 1
    2n = 6
     n = 3

So n=3 and therefore

{{{  sqrt ( (3n-5)/(n+1) )}}} = {{{  sqrt ( (3(3)-5)/((3)+1) )}}} = {{{  sqrt ( (9-5)/(3+1) )}}} = {{{4/4}}} = 1

Edwin</pre>