Question 565387
Call the 3 integers {{{ n }}}, {{{ n+1 }}}, and {{{ n+2 }}}
given:
{{{ n*( n + 1 ) = ( n+2 )^2 - 16 }}}
{{{ n^2 + n = n^2 + 4n + 4 - 16 }}}
Subtract {{{ n^2 }}} from both sides
{{{ n = 4n + 4 - 16 }}}
Subtract {{{ 4n }}} from both sides
{{{ -3n = 4 - 16 }}}
{{{ -3n = -12 }}}
Divide both sides by {{{ -3 }}}
{{{ n = 4 }}}
{{{ n + 1 = 5 }}}
{{{ n + 2 = 6 }}}
The numbers are 4, 5, and 6
check:
{{{ n*( n + 1 ) = ( n+2 )^2 - 16 }}}
{{{ 4*5 = 6^2 - 16 }}}
{{{ 20 = 36 - 16 }}}
{{{ 20 = 20 }}}
OK