Suppose $lcm(a,b)=l$ exist. Then $gcd(\frac{l}{a},\frac{l}{b})=1$. If not then the gcd is $d$.Suppose and $m=\frac{l}{d}<l$,then $m$ would be divisile by both $a$ and $b$ contradictng the fact that $l$ is the LCM. Therefore by Bezout's identity $\exists x_0,y_0 \in \Bbb Z$ such that, $$\frac{l}{a}x_0+\frac{l}{b}y_0=1$$ $$\implies l=\frac{ab}{bx_0+ay_0}=\frac{ab}{n_0*gcd(a,b)},ab\ne 0$$ This suggest $\frac{l}{a}=\frac{b}{n_0*gcd(a,b)}$ and $\frac{l}{b}=\frac{a}{n_0*gcd(a,b)}$ $\implies n_0*gcd(a,b)|a$ and $n_0*gcd(a,b)|b$ $\implies n_0*gcd(a,b)|gcd(a,b) \implies n_0=\pm 1$ making $l=\frac{ab}{gcd(a,b)}$. As this number always exists for $a,b>0$, we conclude $l$ exists and the formula is as above. Please verify if my proof is correct and if there are any gaps in the proof.I doubt the step that $bx_0+ay_0$ is forced to be $gcd(a,b)$, it could be $n_0•gcd(a,b)$ where $n_0> 1 $ this would imply we cannot get an $l$ that has the assumed property of LCM. In other words it is the existence of LCM that i cannot prove as it is only assumed at the start.( if $l$ exists then $n_0=1$ is forced, but $l$ need not exist).
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solution-verificationquestion to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Jun 06 '25 at 15:54