The number of ordered pairs (x, y) of integer | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the equation $x^3+y^3=65$.We can factorize the expression as $(x+y)(x^2-xy+y^2)=65$.Since $x$ and $y$ are integers,$(x+y)$ must be a divisor

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given the equation $x^3+y^3=65$.We can factorize the expression as $(x+y)(x^2-xy+y^2)=65$.Since $x$ and $y$ are integers,$(x+y)$ must be a divisor of $65$.The divisors of $65$ are $\pm 1, \pm 5, \pm 13, \pm 65$.Testing integer values for $x$ and $y$:If $x=1$,then $1+y^3=65 \implies y^3=64 \implies y=4$.If $y=1$,then $x^3+1=65 \implies x^3=64 \implies x=4$.Thus,$(1, 4)$ and $(4, 1)$ are solutions.For other values,if $x$ or $y$ are negative or larger,the sum of cubes does not yield $65$ for integer pairs.Therefore,there are exactly $2$ ordered pairs.

The number of ordered pairs $(x, y)$ of integers satisfying $x^3+y^3=65$ is

Similar Questions

$p$ and $q$ are two roots of the equation $x^2+7x+3=0$. If $\frac{3p}{1-2p}$ and $\frac{3q}{1-2q}$ are the roots of $lx^2+mx+n=0$ and the greatest common divisor of $l, m, n$ is $1$,then $l-m+n=$

If $\left|\frac{x^2+k x+1}{x^2+x+1}\right| < 3$ for all real numbers $x$,then the range of the parameter $k$ is

If $x = \frac{1}{2} \left( \sqrt{7} + \frac{1}{\sqrt{7}} \right)$, then $\frac{\sqrt{x^2 - 1}}{x - \sqrt{x^2 - 1}}$ is equal to

If $\alpha$ is a root of the quadratic equation $x^2 + 6x - 2 = 0$,then another root $\beta$ is

The least positive integer $x$ satisfying $2^{2010} \equiv 3x \pmod{5}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module