Question 894080
n and n+1 are the consecutive integers.


{{{1/(n(n+1))=1/73}}}.


Built this way:
The product of the integers, {{{n(n+1)}}}.

The reciprocal of the product, {{{1/(n(n+1))}}}.


That is {{{1/73}}},
{{{1/(n(n+1))=1/73}}}.


This is equivalent to {{{n^2+n=73}}}, which MUST be factorable using only integers.   <b>73 is a PRIME number.</b>
Discriminant of the equation, {{{n^2+n-73=0}}}.
{{{1^2+4*73=293}}}, which is NOT a perfect square.  This means neither n nor n+1 is an integers.


The question or problem description is wrong.