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\(a,b,c,d \) are integers such that \(ad+bc \) divides each of \(a,b,c,d \) then prove that \(ad+bc = \pm 1 \).

Note by Vilakshan Gupta 11 months, 1 week ago

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First, observe that ad + bc cannot be 0. Now, let the gcd (a, b, c, d) = x. Then,

a = l * x, b = k * x, c = m * x, d = n * x and gcd (l, k, m, n) = 1.

ad + bc = x^2(l * n) + x^2(k * m) = x^2 (l * n + k * m)

ad + bc divides a, b, c, d, and so does x^2 (l * n + k * m). But we have set the gcd (a, b, c, d) to be x. Therefore,

x = x^2 (l * n + k * m), and because (l * n + k * m) is an integer and cannot be 0 (recall the first observation), x has to be 1 or -1.

x = ± 1

(l * n + k * m) = ± 1

ad + bc = ± 1

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## Comments

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TopNewestFirst, observe that ad + bc cannot be 0. Now, let the gcd (a, b, c, d) = x. Then,

a = l * x, b = k * x, c = m * x, d = n * x and gcd (l, k, m, n) = 1.

ad + bc = x^2(l * n) + x^2(k * m) = x^2 (l * n + k * m)

ad + bc divides a, b, c, d, and so does x^2 (l * n + k * m). But we have set the gcd (a, b, c, d) to be x. Therefore,

x = x^2 (l * n + k * m), and because (l * n + k * m) is an integer and cannot be 0 (recall the first observation), x has to be 1 or -1.

x = ± 1

(l * n + k * m) = ± 1

ad + bc = ± 1

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