If m and n are positive integers and (m-n) is | vedclass

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

Question

(A) We know that $m^{2}-n^{2} = (m-n)(m+n)$.Given that $(m-n)$ is an even number,let $(m-n) = 2k$ for some integer $k$.Since $(m+n) = (m-n) + 2n$,and

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) We know that $m^{2}-n^{2} = (m-n)(m+n)$.Given that $(m-n)$ is an even number,let $(m-n) = 2k$ for some integer $k$.Since $(m+n) = (m-n) + 2n$,and $(m-n)$ is even,$(m+n)$ must also be an even number because the sum of two even numbers is even.Let $(m+n) = 2j$ for some integer $j$.Therefore,$(m^{2}-n^{2}) = (2k)(2j) = 4kj$.This shows that $(m^{2}-n^{2})$ is always divisible by $4$.

If $m$ and $n$ are positive integers and $(m-n)$ is an even number,then $(m^{2}-n^{2})$ will be always divisible by

Similar Questions

The largest natural number by which the product of three consecutive even natural numbers is always divisible,is

$(21.5 \% \text{ of } 999)^{1/3} + (42 \% \text{ of } 601)^{1/2} + ? = 28$ (Assuming the equation equals $28$ for a standard evaluation)

If the sum of two numbers is $22$ and the sum of their squares is $404,$ then the product of the numbers is

Three-fourths of one-fifth of a number is $60$. The number is:

$4 \frac{4}{5} \div 6 \frac{2}{5} = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module