Another proof... (division algorithm)?

Question

Theorem:
m and n are two integers where m dne. 0. There exists a unique pair of ints q and r, that satisfy,
n = mq + r and 0 less than equal to r less than abs value of m.

Answer 1

Consider the set of integers n - tr, for all integer values of t. From the least-integer principle, we know this set has a least nonnegative value: this will be the remainder r. You have to show two things:
1) 0 <= m < r
2) r is unique.
See e.g. http://www.mathpath.org/concepts/divisionalgo.htm for details.

Answer 2

Say, n = 2k + a million Then n^2 = 4k^2 + 4k + a million = 4k(ok+a million) + a million Now, for any integer a, the two a is itself even or a is ordinary and then a+a million is even. So, a(a+a million) is often even. to that end, ok(ok+a million) = 2p, say. n^2 = 4*2p + a million = 8p + a million for this reason, n^2 = 2 (mod 8) For the 2d, say the numbers are a, a+a million and a+2 Then the product is, x = a(a+a million)(a+2) Now, as in the previous, a(a+a million) is even. to that end 2 | x returned, for there are 3 opportunities. a = 0(mod 3) => a = 3k a = a million(mod 3) => a = 3k + a million a = 2(mod 3) => a = 3k + 2 If a = 0(mod 3), then 3 | a If a = a million(mod 3), then a + 2 = 3k + 3 = 3(ok+a million) and to that end a+2 = 0(mod 3) => 3 | a+2 further, a = 2(mod 3) => a+a million = 3(mod 3) => a+a million = 0(mod 3) => 3 | a+a million for this reason, the two 3|a or 3|(a+a million) or 3|(a+2) to that end, 3|a(a+a million)(a+2) i.e 3|x Now 2|x and 3|x and 2,3 are primes. for this reason, 2*3 = 6|x.

Answer 3

For a proof, consult any text on Elementary Number Theory. If you have a question about the proof, please specify your difficulty.