Question 862530
m and n, the consecutive integers.

Let {{{m<n}}}.  {{{n=m+1}}}.
{{{15(1/m-1/n)=2}}}------the symbolized translation.
Substitute for n.
{{{15(1/m-1/(m+1))=2}}}
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Solve for m.
{{{15m(m+1)(1/m-1/(m+1))=m(m+1)2}}}
{{{15(m+1)-15m=2m^2+2m}}}
{{{15m+15-15m=2m^2+2m}}}
{{{15=2m^2+2m}}}
{{{2m^2+2m-15=0}}}
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That quadratic is expected to be factorable because you know m and n are integers.
(2m+3)(m-5)????? No.
(2m+5)(m-3)? No.
(2m+15)(m-1)? No.
(2m+1)(m+15)?


Try again.  Integers n and n+1.
{{{15(1/n-1/(n+1))=2}}}, exact translation of the description.
{{{n(n+1)15(1/n-1/(n+1))=2*n(n+1)}}}
{{{15(n+1)-15n=2n^2+2n}}}
{{{15n+15-15n=2n^2+2n}}}
{{{-15=2n^2+2n}}}
{{{2n^2+2n+15=0}}}
MUST be factorable.
(2n+3)(n+5)? No.
(2n+5)(n+3)?  No.
Check the discriminant.  {{{discriminant=2*2-4*2(15)<0}}}.


The nonfactorability in the second attempt and the negative disciminant mean that the question description is wrong.