Question

if x and y are integers then which of the following could be the value of x where x^2-y^2 =121.
a.15
b.33
c.61
d.91

i received one answer that was wrong. I need correct answer along with explanation.
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Answer 1

Before We Solve

An equation with two variables is unsolvable since we cannot eliminate one of them. However, there is a solution since we are given $x,y$ are integers. The sum and difference of two integers are always integers. We are going to use this fact to solve our problem.

Given Equation

$x^2-y^2=121\rightarrow (x+y)(x-y)=11^2$

Solution

The left-hand side is the product of two integers, so each factor will be factors of $11^2$.

The possible pairs of two factors are below.

$(11^2,1)$, $(11,11)$, $(1,11^2)$, $(-11^2,-1)$, $(-11,-11)$, $(-1,-11^2)$

Now equating the factor and the pairs, we obtain,

$\displaystyle{\left \{ {{x=61} \atop {y=60}} \right. }$ or $\displaystyle{\left \{ {{x=11} \atop {y=0}} \right. }$ or $\displaystyle{\left \{ {{x=61} \atop {y=-60}} \right. }$ or $\displaystyle{\left \{ {{x=-61} \atop {y=-60}} \right. }$ or $\displaystyle{\left \{ {{x=-11} \atop {y=0}} \right. }$ or $\displaystyle{\left \{ {{x=-61} \atop {y=60}} \right. }$.

So, option c is correct.

More Information

These are integer points of the graph $y=\pm\sqrt{x^2-121}\:\:(x\leq -11,x\geq 11)$.