Calculus 3 Bars

Question

How many integer solutions are there to the equation x+y+z=8 for which x,y,and z are all greater or equal than −5?  

Answer 1

Lemma:

Assume a given equation for some integers a and n

x + y = n (1)

with the constraints

a ≤ x (2)

a ≤ y (3)

Claim 1: If n < 2a then there are no solutions.

Claim 2: If n ≥ 2a then there are n - 2a + 1 integer solutions.

Proof:

Claim 1 follows from 2a ≤ x + y,

which is the result of adding the inequalities (2) and (3).

To prove Claim 2 note that we get a single solution for any choice of y in the range

a ≤ y ≤ n - a

And there are n - 2a + 1 integers in the range a ≤ y ≤ n - a.

QED

Now consider the given equation

x + y + z = 8 (4)

with the constraints

-5 ≤ x (5)

-5 ≤ y (6)

-5 ≤ z (7)

Any solution satisfies

z = 8 - (x + y) (8)

and

-10 ≤ x + y (9)

(the latter being the result of adding inequalities (5) and (6)).

Combination of (7), (8), and (9) yields

-5 ≤ z ≤ 18 (10)

There are 24 integers z satisfying (10) and for each we can rewrite (4) into

x + y = 8 - z

This is an equation satisfying the prerequisites of the Lemma with a = -5 and n = 8 - z.

By the Lemma, each choice of z gives 8 - z - 2×(-5) + 1 = 19 - z solutions.

The minimal number of solutions is 1, which happens for z = 18,

and the maximal number of solutions is 24, which happens for z = -5.

Using the formula for the sum of arithmetic sequence, the total number of solutions is

1 + 2 + ... + 24 = (1 + 24)×24/2 = 300