Question 219182
Let {{{n}}} be the 1st integer
Then {{{n + 1}}} will be the 2nd integer
given:
{{{sqrt(n) = m}}}
{{{sqrt(n+1) = m+1}}}
Square both sides of both equations
(1){{{n = m^2}}}
{{{n + 1 = (m+1)^2}}}
(2){{{n + 1 = m^2 + 2m + 1}}}
Substitute (1) in (2)
{{{m^2 + 1 = m^2 + 2m + 1}}}
{{{1 = 2m + 1}}}
{{{2m = 0}}}
{{{m = 0}}}
{{{n = m^2}}}
{{{n = 0}}}
{{{n+1 = 1}}}
The numbers are 0 and 1