Question 391464


two consecutive integers are n, n +1
{{{n < n +1}}}

The sum of the squares of two consecutive integers is {{{n^2 +( n +1)^2}}}

{{{n^2 +( n +1)^2 = 8n +9}}}

{{{n^2 + n^2 +2n +1 = 8n +9}}}

{{{2n^2  +2n +1 = 8n +9}}}


{{{2n^2  +2n  -8n = 9 - 1}}}

{{{2n^2  - 6n = 8}}}

{{{2n^2 /2 - 6n/2 = 8/2}}}




{{{n^2  - 3n - 4=0}}}........solve for {{{n}}}

*[invoke quadratic_formula 1, -3, -4, "x"]


{{{x=n}}}

so, you have

{{{n1=4}}}  and 

second integer {{{n1 +1= 4+1=5}}}

4 and 5

{{{n2= -1}}}
second integer {{{n2 +1= -1+1=0}}}

-1 and 0

check: 4 and 5

{{{4^2 +( 5)^2 = 8*4 +9}}}

{{{16 +25 = 32 +9}}}

{{{41 = 41}}}

other pair: -1 and 0

{{{(-1)^2 +( 0)^2 = 8*(-1) +9}}}

{{{1 +0 = -8 +9}}}

{{{1 +0 = 1}}}