<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] I assumed $a$ and $b$ were positive integers.
We have $\mathrm{lcm}(a,a+5) = \frac {a(a+5)}{\gcd(a,a+5)}$, so that the given equality implies
$$\frac {a(a+5)}{\gcd(a,a+5)} = \frac {b(b+5)}{\gcd(b,b+5)}.$$
But $\gcd(x,y) = \gcd(x,y+kx)$ for all $k \in \mathbb{Z}$ so that $\gcd(a,a+5) = \gcd(a,5)$ and $\gcd(b,b+5) = \gcd(b,5)$.
Now if $a$ and $b$ are not multiples of $5$ we have $\gcd(a,5) = 1$ and $\gcd(b,5) = 1$ then
$$\frac {a(a+5)}{\gcd(a,a+5)} = \frac {b(b+5)}{\gcd(b,b+5)} \implies \frac{a*(a+5)}{1} = \frac{b*(b+5)}{1} \implies a = b.$$
If both $a$ and $b$ are multiple of $5$ then $\gcd(a,5) = \gcd(b,5) = 5$ so that 
$$\frac {a(a+5)}{\gcd(a,a+5)} = \frac {b(b+5)}{\gcd(b,b+5)} \implies a(a+5) = b(b+5).$$From here we can conclude that $a=b$ by completing the square both sides and using the fact that a,b are positive.
Edit : 
If $a$ is a multiple of $5$ and not $b$ we have $$a(a+5)=5b(b+5).$$ But $a=5a′$, then the last equality becomes $$5a′(a′+1)=b(b+5).$$ The left side is a multiple of $5$ but not the right side so this case never happens (same reasoning for the case $b$ multiple of $5$ and not $a$).
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