<s>[INST] Prove that if LCMs are equal, then the numbers are equal too. For $a,b \in \mathbb{N}$, how do I prove: 

$$lcm(a,a+5)=lcm(b,b+5) \implies a=b$$

 [/INST] As Batominovski pointed out in his comment, $a$ and $b$ have to be positive integers, otherwise there are counterexamples to the statement.
We're going to use the following identities: 
i) $ab=[a,b](a,b)$ where $[a,b]$ and $(a,b)$ denotes the $\text{lcm}$ and the $\gcd$ of $a$ and $b$, respectively. 
ii) $[ca,cb]=c[a,b]$.
We have $(a,a+5)\mid (a+5)-a=5$, so $(a,a+5)=1$ or $(a,a+5)=5$. Analogously, $(b,b+5)=1$ or $(b,b+5)=5$. So we have four cases to analyze:
I) $(a,a+5)=(b,b+5)=1:$ By the first identity we deduce that $$a(a+5)=[a,a+5]=[b,b+5]=b(b+5).$$ Now, if $a>b$, then $a(a+5)>b(b+5)$ and if $a<b$, then $a(a+5)<b(b+5)$, therefore it must be $a=b$.
II) $(a,a+5)=1$ and $(b,b+5)=5$. Let's write $b=5b'$, then $b+5=5(b'+1)$. On the other hand, by the first identity we have $$a(a+5)=[a,a+5]=[b,b+5]=[5b',5(b'+1)]\overset{\text{by (ii)}}=5[b',b'+1]=5b'(b'+1).$$
Thus, $5\mid a(a+5)$, so $5\mid a$ and then $5\mid a+5$, contradiction; or $5\mid a+5$ and then $5\mid a$, contradiction again.
III) $(a,a+5)=5$ and $(b,b+5)=1$. By the symmetry of $a$ and $b$ this case is analogous to the previous one. 
IV) $(a,a+5)=(b,b+5)=5$. Using the first identity we deduce that $$a(a+5)=5[a,a+5]=5[b,b+5]=b(b+5).$$
So, if $a>b$ or $a<b$ we will have $a(a+5)>b(b+5)$ or $a(a+5)<b(b+5)$, respectively. Hence, it must be $a=b$.
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