To find the number of solutions to the equation $x + y + z + w = 1$, where $x$, $y$, $z$, and $w$ are integers greater than $-4$, we can follow these steps:

1. Transform the Variables: To simplify the problem, let $x' = x + 4$, $y' = y + 4$, $z' = z + 4$, and $w' = w + 4$. With this substitution: $x', y', z', w' \geq 1$ (since $x, y, z, w$ are each greater than $-4$).

2. Rewrite the Equation: Substitute $x = x' - 4$, $y = y' - 4$, $z = z' - 4$, and $w = w' - 4$ into the original equation: $(x' - 4) + (y' - 4) + (z' - 4) + (w' - 4) = 1$ Simplifying this, we get: $x' + y' + z' + w' = 17$

3. Count the Number of Solutions: We now need to count the number of integer solutions to $x' + y' + z' + w' = 17$, where $x', y', z', w' \geq 1$. This is a typical "stars and bars" combinatorial problem with constraints.

4. Remove the Lower Bound: Since $x', y', z', w' \geq 1$, we let $x'' = x' - 1$, $y'' = y' - 1$, $z'' = z' - 1$, and $w'' = w' - 1$, where $x'', y'', z'', w'' \geq 0$. Then: $(x'' + 1) + (y'' + 1) + (z'' + 1) + (w'' + 1) = 17$ Simplifying, we get: $x'' + y'' + z'' + w'' = 13$

5. Final Count: We now need to find the number of non-negative integer solutions to $x'' + y'' + z'' + w'' = 13$, which is given by the formula for combinations with repetition: $\binom{13 + 4 - 1}{4 - 1} = \binom{16}{3}$ Calculating this, we have: $\binom{16}{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$

Thus, there are 560 solutions to the equation $x + y + z + w = 1$ with the given constraints.

Would you like further details or have any questions?

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Related Questions:

1. How would the solution change if the sum was different, such as $x + y + z + w = 5$?
2. What is the approach if only one of the variables is restricted to integers greater than $-4$?
3. Can this solution approach be used if the variables were restricted to positive integers only?
4. What happens if we add an additional variable $v$ to the equation $x + y + z + w + v = 1$?
5. How does the method differ if we impose an upper limit on the values of $x$, $y$, $z$, and $w$?

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Tip:

When dealing with problems involving integer constraints, transforming variables to non-negative values can often simplify combinatorial counting.