Question

Suppose that p is a prime and that x and y are integers with xy≡0(modp). Prove that x≡0(modp) or y≡0(modp).

Answer 1

If (p) is a prime and
`\(xy\equiv 0 \pmod{p}\)`, then either
`\(x\equiv 0 \pmod{p}\) or \(y\equiv 0 \pmod{p}\).`

Step-by-step explanation:

In modular arithmetic,
`\(xy\equiv 0 \pmod{p}\)` implies that (p) divides (xy). Since (p) is a prime number, it must divide at least one of the factors, (x) or (y), due to the fundamental theorem of arithmetic. Without loss of generality, assume (p) divides (x). Then,
`\(x\equiv 0 \pmod{p}\).`

Now, let's delve into the details. If
`\(xy\equiv 0 \pmod{p}\)`, it means (xy) is a multiple of (p). This implies that (p) divides (xy), and since (p) is prime, it must divide either (x) or (y). Without loss of generality, suppose (p) divides (x), meaning (x) is a multiple of (p). Thus,
`\(x\equiv 0 \pmod{p}\).`

In summary, the proof relies on the fact that if a prime number divides the product of two integers, it must divide at least one of the integers individually. In this case, since (p) is prime and (xy) is divisible by (p), (p) must divide either (x) or (y), leading to the conclusion that
`\(x\equiv 0 \pmod{p}\) or \(y\equiv 0 \pmod{p}\).`