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Find all natural numbers $k$ for which there exist natural numbers $n,m$ such that $m(m+k) = n(n+1)$.

Rearranging the given equation gives $k = \dfrac{n(n+1)}{m}-m$. Thus, $m \mid n(n+1)$ and so we have cases.

Case 1: $n = am$ for some $a \in \mathbb{Z}^+$ In this case we have $k = a(am+1)-m$, which is positive for all $a,m$.

Case 2: $n+1= am$ for some $a \in \mathbb{Z}^+$ In this case we have $k = a(am-1)-m$ and we need $k$ to be positive and so $a(am-1)-m \geq 1$, or $am \geq m+1$. This is true if and only if $a \geq 2$.

I didn't see how to solve the other case where both are not multiples of $m$.

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    $$k=\frac{n(n+1)}{m}-m$$ it is necessary to solve this equation? – 2017-01-02
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    @individ Yes, so $m \mid n(n+1)$. – 2017-01-02
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    Also $n(n+1)>m^2$ since $k$ is natural number. – 2017-01-02
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    to factorize and present. $n=am$ ; $n=am-1$ – 2017-01-02
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    @individ Is how I wrote the solutions correct or is there a simpler way? – 2017-01-02
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    Why do you split into cases where $m$ must divide either $n$ or $n+1$? It could contain some factors of $n$ and some of $n+1$, e.g. $n = 8$ and $m = 6$. In this case you find $k = 6$. – 2017-01-02
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    just because $n|m(m+1)$ it doesn't mean it divides one of them. – 2017-01-02
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    @user19405892 What about $6\mid 3\cdot 4$? What Euclid's lemma says is that if $\gcd(a,b)=1$ and $a\mid bc$ then $a\mid c$. That is, the hipothesis is that the divisor and one factor are coprime, not that the factors are coprime. – 2017-01-02
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    @user19405892 Let $n=6$ and $m=2$. – 2017-01-02
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    @JorgeFernándezHidalgo How do I continue from $m \mid n(n+1)$? – 2017-01-02
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    I fon't know reallly, although it looks like you may need to bust out the Pell equation, although hopefully not. – 2017-01-02
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    @ajotatxe I get what you are saying but $n\ge m$ is must. – 2017-01-02

2 Answers 2

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The equation

$$m(m+k) = n(n+1)$$

has positive integer solutions $m,n$ for all positive integers $k$ except $k = 2,3$.

Proof:

If $k = 1$, just let $m = n$, where $n$ is an arbitrary positive integer.

If $k$ is even, $k > 2$, let

$$m = (k(k-2))/4$$

$$n = ((k+2)(k-2))/4$$

and if $k$ is odd, $k > 3$, let

$$m = ((k-1)(k-3))/8$$

$$n = ((k+3)(k-3))/8$$

In both cases, the equation $m(m+k) = n(n+1)$ is identically satisfied.

Next, suppose $k = 2$. Then

$$m(m + 2) = n(n + 1)$$

implies $m < n$ and $m > n-1$, contradiction.

Finally, suppose $k = 3$. Then

$$m(m + 3) = n(n + 1)$$

implies $m < n$ and $m > n -2$, hence $m = n -1$. But then

$$m(m + 3) = n(n+1) \implies (n-1)(n+2) = n(n+1) \implies n = \tfrac{1}{2}$$

contradiction.

This completes the proof.

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    For $ k = 4 $, $ m = k^2 / 2 = 4^2 / 2 = 8 $ and $ n = (k^2 - 2)/2 = 14/2 = 7 $, but $ m(m+4) = 8 \cdot 12 \neq n(n+1) = 7 \cdot 8 $. – 2017-01-02
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    I pasted the wrong formulas -- will edit ... – 2017-01-02
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    All fixed (I think). – 2017-01-02
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For each positive integer $N$, let $\tau(N)$ denote the number of positive integers $d$ that divides $N$. For a given positive integer $u$, we also denote by $\bar{\tau}_u(N)$ the number of odd positive integers $d1$.

We claim that, for a given positive integer $k>1$, the number of solutions $(m,n)\in\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}$ to the Diophantine equation $$m(m+k)=n(n+1)$$ is $$s_k:=\left\{ \begin{array}{ll} \bar{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)+\tilde{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)&\text{if }k\text{ is odd}\,, \\ \dfrac{1}{2}\,\tau(k^2-1)-1&\text{if }k\text{ is even}\,. \end{array}\right.$$ Moreover, we set $s_1:=\infty$. In particular, $s_k=0$ if and only if $k=2$ or $k=3$. For an even integer $k>3$, there exists a unique solution if and only if $(k-1,k+1)$ is a pair of twin primes. For an odd integer $k>3$, there exists a unique solution if and only if $k=7$ or $k=2F-1$, where $F$ is a Fermat prime (i.e., $F$ is a prime such that $F=2^{2^{l}}+1$ for some nonnegative integer $l$).

When $k=1$, there are $s_1=\infty$ solutions to the case $k=1$ given by $m=n$. Suppose from now on that $k>1$. Clearly, we hae $m

If $k$ is odd, we have $$2n\,\left(\frac{k-1}{2}-t\right)=t(k-t)\,.$$ That is, since $t\equiv \frac{k-1}{2}\pmod{\frac{k-1}{2}-t}$, we have $$\dfrac{k-1}{2}-t \,\mid\,\dfrac{k^2-1}{4}\,.$$ Indeed, for any divisor $d\in\mathbb{Z}_{>0}$ of $\dfrac{k^2-1}{4}$ such that $d<\dfrac{k-1}{2}$, we take $t:=\dfrac{k-1}{2}-d$, $$n:=\frac{\left(\frac{k-1}{2}-d\right)\left(\frac{k+1}{2}+d\right)}{2d}\text{ and }m:=n-t=n-\frac{k-1}{2}+d$$ to obtain a solution. Now, note that $n$ defined in this way is a positive integer if and only if $d$ is odd, or $d$ is a maximally even divisor of $\dfrac{n^2-1}{4}$ (that is, $d$ is an even divisor such that $2d$ is not a divisor). There are $s_k=\bar{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)+\tilde{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)$ possible values of $d$.

If $k$ is even, we have $$4n\,(k-2t-1)=(2t)\,(2k-2t)\,.$$ Note that $2t\equiv k-1\pmod{k-2t-1}$, so $$k-2t-1\,\mid\,k^2-1\,.$$ Indeed, let $d\in\mathbb{Z}_{>0}$ be a divisor of $k^2-1$ such that $d

The first odd value of $k>1$ with $s_k>1$ is $k=11$, where we have $s_{11}=3$ solutions $$(m,n)=(1,3)\,,\,\,(m,n)=(3,6)\,,\text{ and }(m,n)=(10,14)\,.$$ The first even value of $k>1$ with $s_k>1$ is $k=8$, where we have $s_8=2$ solutions $$(m,n)=(2,4)\text{ and }(m,n)=(12,15)\,.$$ In both odd and even cases, quasi's solutions correspond to $d=1$.