Question 761695
The sum of the squares of two consecutive integers is thirty one less than twice the square of the second integer. What are the integers? 


I think you got stuck because you have not actually *squared* the numbers in the expressions you wrote.

Solution:

Let the 2 integers be x and x+1.

Square of the 1st integer = {{{x^2}}}
Square of the 2nd integer = {{{(x+1)^2 = x^2 + 2*x + 1}}}

Sum of the squares = {{{x^2 + x^2 + 2*x + 1 = 2*x^2 + 2*x + 1}}} (expression 1)

Twice the square of second integer = {{{2*(x^2 + 2*x + 1) = 2*x^2 + 4*x + 2}}} (expression 2)

Sum of the squares (expression 1) is 31 less than twice the square of the second number (expression 2)

So {{{2*x^2 + 2*x + 1 = 2*x^2 + 4*x + 2- 31}}}

Simplifying and cancelling out the 2*x^2 on both sides

{{{2*x + 1 = 4*x - 29}}}

{{{2*x = 30}}} or

{{{x = 15}}}

The two numbers are 15 and 16.


Sum of squares = 225 + 256 = 481.
Twice square of 16 = 512.
512 - 481 = 31. Check!

:)