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For $a,b \in \mathbb{N}$, how do I prove:

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

user76568
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idliketodothis
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3 Answers3

5

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 $a$).

2

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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Xam
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  • The second link actually goes to the proof of the identity $(ca,cb)=c(a,b)$, but since $\gcd$ and $\text{lcm}$ are dual, the same idea to prove the above identity can be applied to prove the identity ii). – Xam Mar 02 '17 at 19:51
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(1) $a \in \mathbb{N}, p \in \mathbb{P}: p \mid a \iff gcd(a,a+p)=gcd(a,p)=p$

() $a \in \mathbb{N}, p \in \mathbb{P}: gcd(a,p) \neq p \iff gcd(a,p)=1$

"The gcd of a prime $p$ and a natural $a$ is $p$ if it is a divisor of $a$ and $1$ if it is not"

Let $a,b \in \mathbb{N}, p \in \mathbb{P}:$

(2)$$lcm(a,a+p)=\frac{a(a+p)}{gcd(a,p)}=\frac{b(b+p)}{gcd(b,p)}=lcm(b,b+p)$$

$(2) \land a \neq b \implies (p \mid a) \oplus (p \mid b)$.

Choose $p \mid a$ and $p \nmid b$:

$$\mathbb{N} \ni \frac{a(a+p)}{p^2}= \frac{1}{p} \frac{a(a+p)}{p} = \frac{lcm(a,a+p)}{p}=\frac{lcm(b,b+p)}{p}=\frac{b(b+p)}{p} \notin \mathbb{N}$$

Contradiction. $(2) \implies a=b$ (It so happens that $5 \in \mathbb{P}$)

user76568
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