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

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(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

$2$

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(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

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