Let a^2 be an integer. Prove that a^2 is divisible by 5 if and only if a is divi

Question

Let a^2 be an integer. Prove that a^2 is divisible by 5 if and only if a is divisible by 5. Prove that for m elementof N, if m > 2 and x = 1 mod m, y = 2 mod m, then xy = 2 mod m. Prove the following statement by considering three possible cases (values) of x mod 3: There is no integer x such that x^2 = 2 mod 3. Prove that, for all integers a and b, a^2 + b^2 = 0, 1, or 2 mod 4. Deduce that there do not exist integers a and b such that a^2 + b^2 = 1234567. For each n elementof Z, define P_n to be the set P_n = {17n, 17n + 1, 17n + 2..., 17n + 16}. Using the Division Lemma, prove that {P_n}_n elementof z is a partition of Z. (Recall the definition of a partition that appears as Definition 1.7 in the supplementary document. The Division lemma is necessary in the proof, so clearly indicate where you used the lemma in your proof.)

Explanation / Answer

An integer a is divisible by 5 iff q = a/5 is an integer iff a = 5q with an integer q. Now if a is divisible by 5 then a = 5q with an integer q and then a^2 = (5q)^2 = 25q^2 = 5(5q^2 ). So a^2 is also 5 times an integer (the integer 5q^ 2 ), and so a^2 is also divisible by 5. Assume now that a is not divisible by 5. By the previous part a = 5q + r with integer q and r and with 0 r < 5. Had r been 0 we’d have had that a = 5q + 0 = 5q is divisible by 5 contrary to assumption. So r = 1 or r = 2,3,4. In the former case a^2 = (5q + 1)^2 = 5(5q^ 2 + 2q) + 1, which is not divisible by 5

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