Triplets of integers in inequality equations

Question

Question: How many triplets $(x, y, z)$ of integers exist that satisfy the following system of inequalities:

$(x + y - z)^2 < 50$

$(y + z - x)^2 < 140$

$(z + x - y)^2 < 180$

No calculator or other tools allowed. We can remove the squares by taking the roots of each equation but also thus adding an upper and lower bound for each equation

$E_1:-7 \le x + y - z \le 7$

$E_2:-11 \le y + z - x \le 11$

$E_3:-13 \le x + z - y \le 13$

We can perform a total summation to earn a large amount of greater enclosings

Summing $E_1+E_2+E_3$, $\Rightarrow -31 \le x + y + z \le 31$

$E_1+E_2$ $\Rightarrow -9 \le y \le 9$

$E_1+E_3$ $\Rightarrow -10 \le x \le 10$

$E_2+E_3$ $\Rightarrow -12 \le z \le 12$

And this is just as far as i got. We know in one sense that this can be represented lika a 3D volume that is simmilar to a 3D parralelogram but we have no way of proving this. In other words these inequalities define a region in 3D space, and we need to count the integer points inside this region. We can also solve it with coding and have found the correct answer that way but we want a method to prove it by hand. The answer we found

2345 Solutions

Geometric depiction

Any help is appreciated! This problem is from a math competition where teams of four were tasked with solving this and 5 other problems out of which this was the hardest. Only Pen and Paper were allowed, No Calculator. The Organizers said they would post a solution but they have not and we have not gotten a hold of the winning team that solved it.

Answer 1

Let A,B,C the name our equations on the left side and then take the same procedure as the original first step of taking the root on both sides such that

$−7≤(x+y−z)=A≤7$

$−11≤(y+z−x)=B≤11$

$−13≤(x+z−y)=C≤13$

It can now be proved that each combination of x,y,z uniqley defines a combination of A,B,C because they are linearly defined equations. This can be proven in many ways but one example is with matrixes.

The equations are:

1. ( A = x + y - z )
2. ( B = y + z - x )
3. ( C = x + z - y )

These equations form a system of linear equations. To see why the mapping is unique, we check if the equations can be inverted (i.e., if we can uniquely solve for (x), (y), and (z) in terms of (A), (B), and (C)).

Writing the system in matrix form: ((Why is this to large?))

[ \begin{bmatrix}1 & 1 & -1 \\-1 & 1 & 1 \\1 & -1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}=\begin{bmatrix}A \\ B \\ C\end{bmatrix}]

Determinant of the coefficient matrix:

The determinant of the coefficient matrix is according to the rule of Sarrus:

det\begin{matrix} 1 & 1 & -1 \\ -1 & 1 & 1 \\ 1 & -1 & 1 \end{matrix} $= (1*1*1)+(1*1*1)+(-1*-1*-1)-((-1*1*1)+(1*1*-1)+(1*-1*1))=4$

Since the determinant is nonzero , the system is invertible. This means we can uniquely solve for (x), (y), and (z) in terms of (A), (B), and (C), and thus each combination of (x, y, z) uniquely determines (A, B, C).

Now let us observe the pairity of the numbers x,y,z and how it affects the pairity of A,B,C. The possibilities are all even, all odd or two even one odd or two odd one even.

Let us go through the possibilities $Even+Even-Even=Even$ for all A,B,C.

$Odd+Odd-Odd=Odd$ for all A,B,C.

$Even + Even -Odd \equiv Even+Odd-Even \equiv Odd $ for all A,B,C

$Odd+Odd-Even \equiv Odd+Even-Odd \equiv Even $ for all A,B,C

This means that the integers A,B,C will always share the same parity and from this we can construct our cases. A is in the given intervall $−7≤A≤7$ and thus has 8 potential odd integer values that it can be and 7 even numbers it could be. If we expand this to the other values we can count the possible combinations of A,B,C.

For all Even combinations $7*11*13$ possibilities

For all odd Combinations $8*12*14$ possibilities

Total amount of possible combinations A,B,C = $(7*11*13)+(8*12*14)=1001+1344=2345$ And since each of these combinations has a uniqley defined integer tripple x,y,z we have proven that there are 2345 possible solutions to the given system of innequaleties from the start

Answer 2

Added sanity checking section at the end of my answer.

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$E_1:-7 \le x + y - z \le 7$
$E_2:-11 \le y + z - x \le 11$
$E_3:-13 \le x + z - y \le 13$

Let $~n~$ be a variable which denotes $~(x + y + z).~$

Let $~f(n)~$ denote the enumeration of all satisfying ordered triplets $~(x,y,z),~$ as a function of $~n.~$

Then, the desired enumeration is

$$\sum_{n = -\infty}^{+\infty} f(n).$$

So, the problem is reduced to determining the viable range for $~n,~$ and then finding the closed form formula for $~f(n).$

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From the constraints:

• $E_4: ~x + y + z = n.$

• $E_5: ~-7 \leq n-2z \leq 7 \implies $
  $-7 \leq 2z - n \leq 7 \implies $
  $n - 7 \leq 2z \leq n + 7.$

• $E_6:~$ similarly
  $~n - 11 \leq 2x \leq n + 11.$

• $E_7:~$ similarly
  $~n - 13 \leq 2y \leq n + 13.$

Adding $~E_5, ~E_6, ~$ and $~E_7,~$ gives

$$3n - 31 \leq 2(x + y + z) = 2n \leq 3n + 31 \implies $$

$$n - 31 \leq 0 \leq n + 31.$$

Therefore, you must have that

$$-31 \leq n \leq 31.$$

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Stars and Bars theory may now be used to obtain the closed form formula for $~f(n).~$ For Stars and Bars theory, see this article and this article.

To ease the syntax, I am going to let $~n_{xl}, ~n_{xu}, n_{yl}, ~n_{yu}, n_{zl}, ~n_{zu}~$ denote the respective lower and upper bounds on $~x,y,~$ and $~z.$

Then:

• $\displaystyle n_{xl} = \left\lceil \frac{n - 11}{2} ~\right\rceil, ~~n_{xu} = \left\lfloor \frac{n + 11}{2} ~\right\rfloor.$

• $\displaystyle n_{yl} = \left\lceil \frac{n - 13}{2} ~\right\rceil, ~~n_{yu} = \left\lfloor \frac{n + 13}{2} ~\right\rfloor.$

• $\displaystyle n_{zl} = \left\lceil \frac{n - 7}{2} ~\right\rceil, ~~n_{zu} = \left\lfloor \frac{n + 7}{2} ~\right\rfloor.$

Then, $~f(n)~$ equals the enumeration of the number of solutions to

• $x + y + z = n.$

• $x,y,z \in \Bbb{Z}.~$

• The lower bounds for $~x,y,~$ and $~z~$ are $~n_{xl}, ~n_{yl}, ~n_{zl}~$ respectively.

• The upper bounds for $~x,y,~$ and $~z~$ are $~n_{xu}, ~n_{yu}, ~n_{zu}~$ respectively.

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I am going to use this answer as a model for how to attack this problem.

The first step is to adjust the lower and upper bounds for each variable so that the lower bounds are $~0.~$

This changes the enumeration to

• $~x + y + z = n - ( ~n_{xl} + n_{yl} + n_{zl} ~).$

• $~x,y,z \in \Bbb{Z_{\geq 0}}.$

• The upper bounds on $~x,y,z~$ are respectively $~( ~n_{xu} - n_{xl} ~), ~( ~n_{yu} - n_{yl} ~), ~( ~n_{zu} - n_{zl} ~).$

Note
To ease the expression of the formula for $~f(n),~$ I will adopt the convention that when $~a < b,~$ that the expression $~\displaystyle \binom{a}{b}~$ will equal $~0.~$

Here, I am going to break the problem into two cases: $~n~$ is even or $~n~$ is odd.

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$\underline{\text{Case 1:} ~n ~\text{is even}}$

Assume that $~n = 2k.$

$$n_{xl}, ~n_{xu}, ~n_{yl}, ~n_{yu}, ~n_{zl}, ~n_{zu} \\ = \left[k - 5\right], ~\left[k + 5\right], ~\left[k - 6\right], ~\left[k +6\right],\\ \left[k - 3\right], ~\left[k + 3\right] ~~\text{respectively}.$$

• $x + y + z = 14 - k.$
• $x,y,z \in \Bbb{Z_{\geq 0}}.$
• $x \leq 10, ~y \leq 12, ~z \leq 6.$

So, when $~n~$ is even, you must have that

$$~0 \leq 14 - \frac{n}{2} \leq 28 \implies -28 \leq n \leq 28.$$

Within this range for $~n,~$ with $~k = \dfrac{n}{2},~$ you have that

$$f(n) = \binom{16 - k}{2} \\ - \left[ ~\binom{5 - k}{2} + \binom{3 - k}{2} + \binom{9 - k}{2} ~\right] \\ + \left[ ~\binom{-8 - k}{2} + \binom{-2 - k}{2} + \binom{-4 - k}{2} ~\right] \\ - \binom{-15 - k}{2}.$$

So, let $~A~$ denote the set

$$A = \{~n \in \Bbb{Z} ~: ~n ~\text{even}, ~-28 \leq n \leq 28 ~\}.$$

Then, the Case 1 partial sum is

$$\sum_{n \in A} f(n).$$

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$\underline{\text{Case 2:} ~n ~\text{is odd}}$

Assume that $~n = 2k + 1.$

$$n_{xl}, ~n_{xu}, ~n_{yl}, ~n_{yu}, ~n_{zl}, ~n_{zu} \\ = \left[k - 5\right], ~\left[k + 6\right], ~\left[k - 6\right], ~\left[k +7\right],\\ \left[k - 3\right], ~\left[k + 4\right] ~~\text{respectively}.$$

• $x + y + z = 15 - k.$
• $x,y,z \in \Bbb{Z_{\geq 0}}.$
• $x \leq 11, ~y \leq 13, ~z \leq 7.$

So, when $~n~$ is odd, with $~k = \dfrac{n - 1}{2},~$ you must have that

$$~0 \leq 15 - \frac{n - 1}{2} \leq 31 \implies -31 \leq n \leq 31.$$

Within this range for $~n,~$ you have that

$$f(n) = \binom{17 - k}{2} \\ - \left[ ~\binom{5 - k}{2} + \binom{3 - k}{2} + \binom{9 - k}{2} ~\right] \\ + \left[ ~\binom{-9 - k}{2} + \binom{-3 - k}{2} + \binom{-5 - k}{2} ~\right] \\ - \binom{-17 - k}{2}.$$

So, let $~B~$ denote the set

$$B = \{~n \in \Bbb{Z} ~: ~n ~\text{odd}, ~-31 \leq n \leq 31 ~\}.$$

Then, the Case 2 partial sum is

$$\sum_{n \in B} f(n).$$

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$\underline{\text{Sanity Checking}}$

Although the Case 1 and Case 2 formulas can be manually applied with scratch paper, it is very tedious to do so.

I wrote a computer program to perform the Case 1 and Case 2 computations, with $~n~$ in the specified Case 1 and Case 2 ranges.

Remembering the convention that $~a < b \implies \displaystyle \binom{a}{b} = 0,~$ the computer program's final computation, $~2345,~$ agrees with the original poster's official solution.