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Question Number 122646 by mathocean1 last updated on 18/Nov/20

Determinate the couples   (a; b) such that   GCD(a;b)+LCM(a;b)=a+b  GCD: greatest common divisor  LCM: least common multiple

$${Determinate}\:{the}\:{couples}\: \\ $$$$\left({a};\:{b}\right)\:{such}\:{that}\: \\ $$$${GCD}\left({a};{b}\right)+{LCM}\left({a};{b}\right)={a}+{b} \\ $$$${GCD}:\:{greatest}\:{common}\:{divisor} \\ $$$${LCM}:\:{least}\:{common}\:{multiple} \\ $$

Commented by mathocean1 last updated on 18/Nov/20

so we have m couples sir?

$${so}\:{we}\:{have}\:{m}\:{couples}\:{sir}? \\ $$

Commented by mr W last updated on 18/Nov/20

a=p_1 p_2 ...p_m  with p=any prime number  b=p_1 ^n_1  p_2 ^n_2  ...p_m ^n_m    gcd(a,b)=a  lcm(a,b)=b

$${a}={p}_{\mathrm{1}} {p}_{\mathrm{2}} ...{p}_{{m}} \:{with}\:{p}={any}\:{prime}\:{number} \\ $$$${b}={p}_{\mathrm{1}} ^{{n}_{\mathrm{1}} } {p}_{\mathrm{2}} ^{{n}_{\mathrm{2}} } ...{p}_{{m}} ^{{n}_{{m}} } \\ $$$${gcd}\left({a},{b}\right)={a} \\ $$$${lcm}\left({a},{b}\right)={b} \\ $$

Commented by mr W last updated on 18/Nov/20

infinite couples!

$${infinite}\:{couples}! \\ $$

Commented by mathocean1 last updated on 18/Nov/20

ok sir

$${ok}\:{sir} \\ $$$$ \\ $$

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