Problem 5

Number 5

If $x$ and $y$ are integers satisfying $x^2 + y^2 = 169$ , then the possible values for $x + y$ are

$\pm 7$
$\pm 10$
$\pm 13$
$\pm 14$

$(1)$ , $(2)$ , and $(3)$ ONLY are correct.

$(1)$ and $(3)$ ONLY are correct.

$(2)$ and $(4)$ ONLY are correct.

ONLY $(4)$ is correct.

ALL choices are correct.

Explanation

Given the equation $x^2 + y^2 = 169$ where $x$ and $y$ are integers. The value $169$ is the square of $13$ ( $13^2 = 169$ ). The integer pairs $(x, y)$ satisfying this circle equation can be found using Pythagorean triples or properties of perfect squares.

The possible pairs are:

If one variable is $0$ , then the other is $\pm 13$ . Pairs: $(13, 0), (-13, 0), (0, 13), (0, -13)$ . Possible values for $x + y$ :

$\pm 13 + 0 = \pm 13$
If both variables are non-zero, we look for combinations of squares summing to $169$ . The Pythagorean triple involving $13$ as the hypotenuse is $(5, 12, 13)$ because $5^2 + 12^2 = 25 + 144 = 169$ . Possible pairs $(x, y)$ (including negative values): $(\pm 5, \pm 12)$ and $(\pm 12, \pm 5)$ . Possible values for $x + y$ from these combinations:

$\begin{aligned} 5 + 12 &= 17 \\ 5 + (-12) &= -7 \\ -5 + 12 &= 7 \\ -5 + (-12) &= -17 \end{aligned}$

Thus, $x + y$ can be $\pm 17$ or $\pm 7$ .