Question 1109206
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The statement of the problem implies that the numbers are whole numbers.  So solving the problem requires less work if you look for numbers for which the difference of the squares is 85 and find such a pair with one of the numbers being 1 less than twice the other.<br>
{{{a^2-b^2 = 85}}}
{{{(a+b)(a-b) = 85}}}<br>
There are two possibilities in whole numbers:
{{{a+b=85}}}; {{{a-b=1}}} --> (a,b) = (43,42)  no....
{{{a+b=17}}}; {{{a-b=5}}} --> (a,b) = (11,6)  YES!!!