lulzmort

Question : for any two numbers p,q which are odd will it always be true that p^2 - q^2 will always be a multiple of 4?

Examples :

13^2 - 5^2 = 169 - 25 = 144 which 144= 36*4

21^2 - 3^2 = 441 - 9 = 432 which 432 = 108*4

7^2 - 11^2 = 49 - 121 = -72 which -72 = -18*4

Proof :

You can construct both p and q such that p = 2a + 1 and q = 2b + 1, where a,b ∈ ℤ

p^2 - q^2

= (p - q)(p + q)

= ((2a + 1) - (2b + 1))((2a + 1)+(2b + 1))

= (2(a-b))(2(a+b) + 2)

= (4(a-b)(a+b) + 4(a-b))

= (4(a-b))(a+b+1)

The 4 in the final product it proves that the difference will have be a multiple of 4.

Sorry if there is any mistakes or convention breaking! Also this looks fairly ugly so if

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