If m, n are any two odd positive integers with n

Question

If m, n are any two odd positive integers with n < m, then the largest positive integer which divides all the numbers of the type m2 - n2 is (A) 4 (B) 6 (C) 8 (D) 9

Tardigrade Admin · Accepted Answer

Let m = 2k + 1, n = 2k - 1 (k ∈ N) ∴ m2 - n2 = 4k2 + 1 + 4k - 4k2 + 4k - 1 = 8k Hence, all the numbers of the form m2 - n2 are always divisible by 8.

Q. If $m, n$ are any two odd positive integers with $n < m$ , then the largest positive integer which divides all the numbers of the type $m^{2} - n^{2}$ is

$4$

$6$

$8$

$9$

Let $m = 2 k + 1, n = 2 k - 1 (k \in N)$
$∴ m^{2} - n^{2} = 4 k^{2} + 1 + 4 k - 4 k^{2} + 4 k - 1 = 8 k$
Hence, all the numbers of the form $m^{2} - n^{2}$ are always divisible by $8$ .

Q. If m,n are any two odd positive integers with n<m, then the largest positive integer which divides all the numbers of the type m2−n2 is

4

6

8

9

Let m=2k+1,n=2k−1(k∈N) ∴m2−n2=4k2+1+4k−4k2+4k−1=8k Hence, all the numbers of the form m2−n2 are always divisible by 8.

Q. If $m, n$ are any two odd positive integers with $n < m$ , then the largest positive integer which divides all the numbers of the type $m^{2} - n^{2}$ is

$4$

$6$

$8$

$9$

Let $m = 2 k + 1, n = 2 k - 1 (k \in N)$
$∴ m^{2} - n^{2} = 4 k^{2} + 1 + 4 k - 4 k^{2} + 4 k - 1 = 8 k$
Hence, all the numbers of the form $m^{2} - n^{2}$ are always divisible by $8$ .