Prove that

(a) there exist integers m and n such that 2m + 7n = 1.

(b) there exist integers m and n such that 15m + 12n = 3.

(c) there do not exist integers m and n such that 2m + 4n = 7.

(d) there do not exist integers m and n such that 12m + 15n = 1.

(e) for every integer t, if there exist integers m and n such that 15m + 16n = t, then there exist integers rand s such that 3r + 8s = t.

(f) if there exist integers m and n such that 12m + 15n = 1, then m and n are both positive.

(g) for every odd integer m, if m has the form 4k + 1 for some integer k, then m + 2 has the form 4j - 1 for some integer j. (h) for every integer m, if m is odd, then m2 = 8k + 1 for some integer k.

(i) for all odd integers m and n, if mn = 4k - 1 for some integer k, then m or n is of the form 4} - 1 for some integer j.

LStep-by-step explanation:

(a) To proof this statement choose m = 4 and n = -1, from here we have 2(4) + 7(-1) = 1 as required. Hence th statement is true.

(b) To proof this statement choose m = 1 and n = -1, from here we have 15(1) + 12(-1) = 3 as required. Hence the statement is true.

(c) To proof this statement, suppose m and n are integers, since integers are closed under addition and multiplication then 2m, 4n and 2m + 4n are all integers. Obviously 2m and 4n are even integers, therefore their sum is also an even integer. This contradicts the equation 2m + 4m = 7, hence there do not exist integers m and n that satisfy the statement.

(d) To proof this statement, suppose m and n are integers, then 12m, 15n and 12m + 15n are all integers, since integers are closed under addition and multiplication. Factoring out 3 from the sum, we have: 3(4m + 5n). Therefore 3(4m + 5n) = 1 is a contradiction, because it will give 4m + 5n = 1/3 and addition of two integers can never give a fraction.

(e) To proof this statement, suppose m and n are integers, then 15m + 16n = t can be rewritten as 3(5m) + 8(2n) = t. Since integers are closed under multiplication, then 5m and 2n are integers. Let 5m = r and 2n = s, then we have: 3r + 8s = t as required.

(f) To proof this statement, suppose that m and n are both negative integers, we see that 12m + 15n = 1 is clearly impossible because the sum would be less than 0. Therefore, if there exist integers m and n such that 12m + 15n = 1, then they must be positive.

(g) To proof this statement, suppose m is an integer and has the form 4k + 1 for some integer k, then m + 2 = 4k + 1 + 2 = 4k + 3 = 4k + 4 - 1 = 4(k + 1) - 1. Since k is an integer then k + 1 is an integer because integers are closed under addition. Therefore, put j = k + 1. Hence we have 4j - 1.

(h) suppose that m is odd, then there is an n such that m = 2n + 1, therefore m² = (2n + 1)² = 4n² + 4n + 1 = 4(n² + n) + 1. Here, we need a lemma that says: suppose k is an integer, k(k + 1) is an even integer. This can be proven intuitively, when k is odd, the term in the bracket becomes even, and the whole expression is even. If k is even, then automatically the whole expression becomes even. Hence, our lemma is proven intuitively. This implies that n² + n is an even integer and can be replaced by 2k for some k that is an integer. Therefore, we have 4(2k) + 1 = 8k + 1 as required.

(i) suppose that m and n are odd integers such that mn = 4k - 1 for some k that is an integer. If we assume that neither m nor is in the for 4j - 1, then:

mn = (4a + 1)(4b + 1) for some a and b that are integers. Expanding that we have mn = 4(4ab + a + b) + 1 which is not in the form 4k - 1 for some k that is an integer as assumed. Therefore, if m and n are odd integers such that mn is in the form 4k - 1 for some k that is an integer, then m and n are in the form 4j - 1 for some j that is an integer.

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