Question 1162750
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The following original response left out important parts of the solution -- though the final answer to the question was correct.  See further down for the corrected/expanded response.<br>
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If a and b are integers and {{{sqrt(a)-sqrt(b) = sqrt(11)}}}, then for some positive integer value of n,<br>
{{{a=(n+1)^2}}} and {{{b = n^2}}}<br>
Then {{{a/b = (n+1)^2/n^2 = ((n+1)/n)^2}}}<br>
For positive integer values of n, the maximum value of a/b is clearly when n=1 and a/b = (2/1)^2 = 4.<br>
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Corrected response....<br>
If a and b are integers and {{{sqrt(a)-sqrt(b) = sqrt(11)}}}, then for some positive integer value of n,<br>
{{{a=(11(n+1)^2)}}} and {{{b = 11(n^2)}}}<br>
With those values of a and b,
{{{sqrt(a) = (n+1)*sqrt(11)}}}
{{{sqrt(b) = (n)*sqrt(11)}}}
{{{sqrt(a)-sqrt(b) = (n+1)*sqrt(11)-(n)*sqrt(11) = sqrt(11)}}}<br>
Then {{{a/b = (n+1)^2/n^2 = ((n+1)/n)^2 = (1+1/n)^2}}}<br>
For positive integer values of n, the maximum value of a/b is clearly when n=1 and a/b = (2/1)^2 = 4.<br>
CHECK....<br>
n=1: a = 11(2^2) = 44; b = 11(1^2) = 11; {{{sqrt(a)-sqrt(b) = 2*sqrt(11)-1*sqrt(11) = sqrt(11)}}}; a/b = 44/11 = 4<br>
n=2: a = 11(3^2) = 99; b = 11(2^2) = 44; {{{sqrt(a)-sqrt(b) = 3*sqrt(11)-2*sqrt(11) = sqrt(11)}}}; a/b = 99/44 = 9/4<br>
n=3: a = 11(4^2) = 176; b = 11(3^2) = 99; {{{sqrt(a)-sqrt(b) = 4*sqrt(11)-3*sqrt(11) = sqrt(11)}}}; a/b = 176/99 = 16/9<br><br>
Clearly for larger values of n the ratio a/b will continue to get smaller....<br>