If $a \mid m$ and $(a + 1) \mid m$, prove $a(a + 1) | m$.

Can anyone help me out here? Can’t seem to find the right rules of divisibility to show this:

If $a \mid m$ and $(a + 1) \mid m$, then $a(a + 1) \mid m$.

Solutions Collecting From Web of "If $a \mid m$ and $(a + 1) \mid m$, prove $a(a + 1) | m$."

It is not surprising that you are finding this difficult, because it goes beyond basic divisibility rules — it rather requires something which is essentially equivalent to the uniqueness of prime factorization. [Edit: Actually this is comment is incorrect — as Robin Chapman’s answer shows, it is possible to prove this using just divisibility rules. In particular it is true in any integral domain.]

I assume $a$ and $m$ are positive integers. The first observation is that $a$ and $a+1$ are relatively prime: i.e., there is no integer $d > 1$ — or equivalently, no prime number– which divides both $a$ and $a+1$, for then $d$ would have to divide $(a+1) – a = 1$, so $d = 1$.

Now the key step: since $a$ divides $m$, we may write $m = aM$ for some positive integer $M$. So $a+1$ divides $aM$ and is relatively prime to $a$. I claim that this implies $a+1$ divides $M$. Assuming this, we have $M = (a+1)N$, say, so altogether

$m = aM = a(a+1)N$, so $a(a+1)$ divides $m$.

The claim is a special case of:

(Generalized) Euclid’s Lemma: Let $a,b,c$ be positive integers. Suppose $a$ divides $bc$ and $a$ is relatively prime to $b$. Then $a$ divides $c$.

A formal proof of this requires some work! See for instance


In particular, proving this is essentialy as hard as proving the fundamental theorem of arithmetic.

The other answers put this in a general context, but in this example one
can be absolutely explicit. If $a\mid m$ and $(a+1)\mid m$ then there are
integers $r$ and $s$ such that
As $r-s$ is an integer, then $a(a+1)\mid m$.

${\bf Proof}\rm\quad\displaystyle \frac{m}{a},\; \frac{m}{a+1}\in\mathbb{Z} \ \,\Rightarrow\,\ \frac{m}{a} – \frac{m}{a\!+\!1} \; = \;\frac{m}{a(a\!+\!1)} \in \mathbb Z\quad$ QED

${\bf Remark}\rm\ \, \text{More generally, if }\, n = bc \,-\, ad \;$ is a linear combination of $\rm\, a, b\, $ then

$\rm\text{we have}\quad\,\ \displaystyle \frac{m}{a},\; \frac{m}{b}\in\mathbb{Z} \;\;\Rightarrow\;\; \frac{m}{a}\frac{bc}{b} – \frac{ad}{a}\frac{m}{b} = \frac{mn}{ab} \in \mathbb Z$

By Bezout, $\rm\, n = gcd(a,b)\, $ is the least positive linear combination, so the above yields

$\rm\qquad\qquad a,b\mid m \;\Rightarrow\; ab\mid m\;gcd(a,b) \;\Rightarrow\; \mathfrak{m}_{a,b}\!\mid m\;\;\;$ for $\;\;\rm \mathfrak{m}_{a,b} := \dfrac{ab}{gcd(a,b)}$

i.e. $ $ every common multiple $\rm\, m\,$ of $\,\rm a,b\,$ is a multiple of $\;\rm \mathfrak{m}_{a,b}. \;$ But $\rm\,\mathfrak{m}_{a,b}\,$ is also a common multiple: $\rm\ a,b\mid \mathfrak{m}_{a,b} \;$ viz. $\displaystyle \,\rm \frac{\mathfrak{m}_{a,b}}{a} = \;\frac{a}{a}\frac{b}{gcd(a,b)}\in\mathbb Z\,$ $\,\Rightarrow\,$ $\rm\, a\mid \frak{m}_{a,b}.\,$ $\rm\, b\mid \mathfrak{m}_{a,b}\,$ by symmetry. Thus $\,\rm \mathfrak{m}_{a,b} = lcm(a,b)\,$ since it’s a common multiple of $\,\rm a,b\,$ that’s divisibility-least, i.e. divides every common multiple. This is the general definition of LCM in an arbitrary domain (ring without zero-divisors), i.e. we have the following universal dual definitions of LCM and GCD:

Definition of LCM $\ \ $ If $\quad\rm a,b\mid c \;\iff\; [a,b]\mid c \quad$ then $\,\quad\rm [a,b] \;$ is an LCM of $\;\rm a,b$

Definition of GCD $\ \ $ If $\quad\rm c\mid a,b \;\iff\; c\mid (a,b) \quad$ then $\quad\rm (a,b) \;$ is an GCD of $\;\rm a,b$

Note that $\;\rm a,b\mid [a,b] \;$ follows by putting $\;\rm c = [a,b] \;$ in the definition. $ $ Dually $\;\rm (a,b)\mid a,b \;$

Such $\iff$ definitions enable slick unified proofs of both arrow directions, e.g.

Theorem $\rm\;\; (a,b) = ab/[a,b] \;\;$ if $\;\rm\ [a,b] \;$ exists.

Proof: $\rm\quad d\mid a,b \iff a,b\mid ab/d \iff [a,b]\mid ab/d \iff\ d\mid ab/[a,b] \quad$ QED

The conciseness of the proof arises by exploiting to the hilt the $\iff$ definition of LCM, GCD. Implicit in the above proof is an innate cofactor duality. Brought to the fore, it clarifies LCM, GCD duality (analogous to DeMorgan’s Laws), e.g. see here and here.

By the theorem, GCDs exist if LCMs exist. But common multiples clearly comprise an ideal, being closed under subtraction and multiplication by any ring element. Hence in a PID the generator of an ideal of common multiples is clearly an LCM. In Euclidean domains this can be proved directly by a simple descent, e.g. in $\:\mathbb Z \;$ we have the following high-school level proof of the existence of LCMs (and, hence, of GCDs), after noting the set $\rm M$ of common multiples of $\rm a,b$ is closed under subtraction and contains $\:\rm ab \ne 0\:$:

Lemma $\ $ If $\;\rm M\subset\mathbb Z \;$ is closed under subtraction and $\rm M$ contains a nonzero element $\rm\,k,\,$ then $\rm M \:$ has a positive element and the least such positive element of $\;\rm M$ divides every element.

Proof $\, $ Note $\rm\, k-k = 0\in M\,\Rightarrow\, 0-k = -k\in M, \;$ therefore $\rm M$ contains a positive element. Let $\rm\, m\,$ be the least positive element in $\rm\, M.\,$ Since $\,\rm m\mid n \iff m\mid -n, \;$ if some $\rm\, n\in M\,$ is not divisible by $\,\rm m\,$ then we
may assume that $\,\rm n > 0,\,$ and the least such. Then $\rm\,M\,$ contains $\rm\, n-m > 0\,$ also not divisible by $\rm m,\,$ and smaller than $\rm n$, contra leastness of $\,\rm n.\ \ $ QED

An alternate route: We show that, if $ax+by=1$ and $m$ is divisible by $a$ and $b$ then $m$ is divisible by $ab$. (Then apply this with $b=a+1$, $x=-1$ and $y=1$.)

Proof: Let $m=ak=bl$. Then $ab(xl+ky)=(ax+by)m=m$. QED

The point here is that the hypothesis $\exists_{x,y}: ax+by=1$ is often easier to use than $GCD(a,b)=1$. The equivalence between these two is basically equivalent to unique factorization, and you can often dodge unique factorization by figuring out which of these two you really need.