what is the equation?

Question

one positive integer is three lager than another positive integer. the sum of the squares of these integer is less than three times the product of the integers. find these integers.

Answer 1

let one integer be n, then the other is n+3..1)

and sum of squares is = n^2 +(n+3)^2
= n^2 +n^2 +6n +9
2n^2 +6n + 9.....2)

now the product is n(n+3)=n^2+3n
=>3 times the product => 3n^2 + 9n ....3)

since both are positive integers hence their sqrs and positive integer multipliers are also positive.

given: 2) is less than 3)

2n^2 +6n+ 9 < 3n^2 +9n

subtract 3n^2 +9n from both sides:
-n^2 -3n +9 < 0
multiply both sides with -1 and this changes the sign of equality:
n^2 +3n -9 > 0
(n-n1)(n-n2)>0
which implies n-n1>0 and n-n2>0
or n>n1 or n2


take the extreme case of equality:
n^2 +3n -9 =0
solve for the roots , n1 and n2 ,of n:
= [-3+- sqrt {45}]/2
=[-3 +- 3sqrt5]/2
discard the negative result as n is supposed to be a positive int.
n = 3{1+sqrt5]/2
=4.85
since n>4.85 hence nearest positive integer is 5 and the other number is 8.
you can counter check by substituting in 2) and3)
[8^2 + 5^2 = 64 +25 = 89 ]< [3( 8*5)= 120]

Answer 2

let x = smaller integer
let y = larger integer
Then y =x+3
x^2+y^2 < 3xy
Substitute:
x^2 + (x+3)^2 < 3x(x+3)
x^2 + x^2 + 6x +9 < 3x^2 + 9x
x^2 +3x > 9
x= 2 is the smallest integer meeting this requirement
So y = x+3 = 2+3=5

Answer 3

equation 1: x+3 = y

equation 2: x^2+y^2 = -3(xy)

two equations, two unknowns...solve for x and y