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  "body": "\n## Proof\n\n1. Let $p$ be a prime and let $a$, $b$ be nonzero integers. Put $\\alpha = \\nu_p(a)$ and $\\beta = \\nu_p(b)$, where $\\nu_p$ is the [[p-adic valuation|04VCPS30ZE]].\n2. Since $p^\\alpha \\mid a$, write $a = p^\\alpha a'$ with an integer $a'$. Then $a' \\neq 0$.\n3. We have $p \\nmid a'$. Otherwise $a' = p c$ for an integer $c$, so $a = p^{\\alpha + 1} c$ and $p^{\\alpha+1} \\mid a$. This contradicts the maximality of $\\alpha$.\n4. In the same way, $b = p^\\beta b'$ with an integer $b'$ and $p \\nmid b'$.\n5. Then $a b = p^{\\alpha + \\beta} a' b'$. Since $a b \\neq 0$, $\\nu_p(ab)$ is a natural number. Since $p^{\\alpha + \\beta} \\mid ab$, $\\nu_p(ab) \\ge \\alpha + \\beta$.\n6. Suppose $p^{\\alpha + \\beta + 1} \\mid ab$. Then $p^{\\alpha + \\beta} a' b' = p^{\\alpha + \\beta + 1} d$ for an integer $d$.\n7. Cancelling the nonzero factor $p^{\\alpha + \\beta}$ gives $a' b' = p d$, so $p \\mid a' b'$.\n8. By [[Euclid's lemma|051VYKA1EX]], $p \\mid a'$ or $p \\mid b'$. This contradicts steps 3 and 4.\n9. So $p^{\\alpha + \\beta + 1} \\nmid ab$. If $k \u003e \\alpha + \\beta$ and $p^k \\mid ab$, then $p^{\\alpha + \\beta + 1} \\mid ab$, since $p^{\\alpha+\\beta+1}$ [[divides|096PW7Q014]] $p^k$. Hence no such $k$ exists.\n10. Therefore $\\nu_p(ab) = \\alpha + \\beta = \\nu_p(a) + \\nu_p(b)$.\n\n## Idea\n\nPull out of $a$ and $b$ all the factors $p$ they contain. The parts that remain are not divisible by $p$, and by Euclid's lemma neither is their product. So the product contains exactly $\\alpha + \\beta$ factors $p$.\n",
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