Let 'm' and 'n' be any two odd numbers, with n less than m.

Question

Let 'm' and 'n' be any two odd numbers, with n less than m. The largest integer which divides all possible numbers of the form  m^2  -  n^2  is:

A. 2
B. 4
C. 6
D. 8
E. 16

IanStewart · Accepted Answer

If m and n are both odd, then m + n is even, and m - n is even, so (m + n)(m - n) will be divisible by 4 for sure. If m - n is divisible by 4, then (m + n)(m - n) will be divisible by 8. If m - n is not divisible by 4, then since m - n is even, it must equal 2*(some odd number). But m - n and m + n are 2n apart, and n is odd, so m + n would then equal (m - n) + 2n = 2*(some odd number) + 2(some other odd number) = 2(odd + odd) = 2*even, and m + n would then need to be divisible by 4. So one of m-n or m+n is always divisible by 4 in this situation, and the other is divisible by 2, so their product is divisible by 8 for sure.

We can't be sure (m + n)(m - n) is divisible by anything larger than 8, because we might have m = 3 and n = 1, in which case m^2 - n^2 = 8.

We can't be sure (m + n)(m - n) is divisible by anything larger than 8, because we might have m = 3 and n = 1, in which case m^2 - n^2 = 8.

sujoykrdatta · Answer

Let the odd numbers be: 
m = 2p+1
n = 2q+1

m² - n² 
= (2p+1)² - (2q+1)²
= (2p+1 -2q-1)(2p+1 + 2q+1)
= 2(p-q)2(p+q+1)
= 4(p-q)(p+q+1)

Possible cases:

A) p and q are both odd or both even: (p-q) is even 
Hence , m²-n² is a multiple of 4*2= 8

B) p is odd and q is even or vice-versa: (p+q+1) is even 
Hence , m²-n² is a multiple of 4*2= 8

Thus, m²-n² is always divisible by 8

Answer D

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alrs_14 · Answer

There is a much faster way of solving the problem. Start by taking 5^2 - 3^2, which is 16, hence, it could be divisible by 16. Now take 7^2 - 5^2, which is 24, which is not divisible by 16, but notice there is a difference of 8. Both numbers are divisible by 8. We are creating a sequence where 16, 24, 32 are all divisible by 8, hence 8 is the right answer.

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Regor60 · Answer

To find the largest number that will divide ALL forms suggests a strategy of making the difference of squares as SMALL as possible, since we will be assured that any difference of squares greater than that minimum will always be divisible by that number.

So, if:

m = 2K+1 and n = 2L+1

To minimize m^2-n^2 we would ideally like them to be the same number, but the best that can be done is to make:

K = L+1

So m = 2L+3 and n = 2L+1

To minimize both, set L=0:

m=3 and n=1

m^2-n^2 = 9-1 = 8

minimizes the difference of squares and represents the largest number dividing all possibilities

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Let 'm' and 'n' be any two odd numbers, with n less than m.

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Let 'm' and 'n' be any two odd numbers, with n less than m.

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