## Task 3A - 261233A

Determine whether there exist four consecutive natural numbers that have the following property:

Each of the four numbers can be decomposed into two positive integer summands $x$ and $y$ such that each number is a divisor of $x \cdot y$.
There are four such numbers. To prove this, it suffices to give an example. Such an example is formed by the numbers 242, 243, 244, 245:

$242=22+220$ is a divisor of $22 \cdot 220$ because $242 \cdot 20=22 \cdot 220$

$243=81+162$ is a divisor of $81 \cdot 162$ because $243 \cdot 54=81 \cdot 162$

$244=122+122$ is a divisor of $122 \cdot 122$ because $244 \cdot 61=122 \cdot 122$

$245=35+210$ is a divisor of $35 \cdot 210$ because $245 \cdot 30=35 \cdot 210$

Heuristic solution:

A natural number $n$ has the mentioned property if and only if ($n \geq 2$ and) $n$ is a divisor of one of the products $k \cdot(n-k)(k=1, \ldots, n-1)$. Since $k \cdot(n-k)=k n-k^{2}$, this is equivalent to $n$ being a divisor of one of the squares $k^{2}(k=1, \ldots, n-1)$.

For this, it is sufficient that the number $n$ is divisible by a square $q^{2}>1$; because if this is the case, there exists a natural number $a$ with $n=q^{2} \cdot a$, and thus $n$ is a divisor of the square of the natural number $k=q a$, which is smaller than $n$ because $k=\frac{n}{q}$ and $q>1$.

Now one can try to find four numbers of the required type, for example, as

$$
\begin{aligned}
n & =2^{2} \cdot a \\
n+1 & =3^{2} \cdot b \\
n+2 & =5^{2} \cdot c \\
n+3 & =7^{2} \cdot d
\end{aligned}
$$

(1) and (2), i.e., $4 a+1=9 b$, can be solved, for example, by $b=1+4 t, a=2+9 t$

$$
n=8+36 t
$$

then (5) and (3), i.e., $10+36 t=25 c$, can be solved, for example, by $t=-10+25 u, c=-14+36 u$

$$
n=-352+900 u
$$

finally, (6) and (4), i.e., $-349+900 u=49 d$, can be solved, for example, by $u=-16+49 v, d=-301+900 v$,

$$
n=-14572+44100 v
$$

for $v=1$ thus $n=29348$.

If the solution is formulated as a proof of sufficient conditions, as here, a check is not necessary.

Otherwise, it is necessary for the correctness of the solution (as noted above, even alone) to confirm the required property:

$29348=4 \cdot 7337=x+y$ is a divisor of $x y$ for $x=2 \cdot 7337, y=2 \cdot 7337$,

$29349=9 \cdot 3261=x+y$ is a divisor of $x y$ for $x=3 \cdot 3261, y=6 \cdot 3261$,

$29350=25 \cdot 1174=x+y$ is a divisor of $x y$ for $x=5 \cdot 1174, y=20 \cdot 1174$,

$29351=49 \cdot 599=x+y$ is a divisor of $x y$ for $x=7 \cdot 599, y=42 \cdot 599$.

Adapted from $[5]$

2. Solution:

First, we consider the following lemma, where only the first implication is relevant for the problem:

A natural number $n$ can be represented as the sum of two positive integers $x$ and $y$ with $n \mid x \cdot y$ if and only if there is a prime number $p$ such that $p^{2}$ is a divisor of $n$.

Proof: First, assume that $n$ is divisible by $p^{2}$ and choose $x:=\frac{n}{p}$ and $y:=n-x$. Then $00$ and $n|x, n| y$, so $n \leq x, y$, which would lead to a contradiction with $x+y=n$. Therefore, there is at least one prime number $p$ with $p^{2} \mid n, \square$.

Now to the problem:

Since $4=2^{2}, 9=3^{2}, 25=5^{2}$, and $49=7^{2}$ are pairwise coprime, there exists (and thus infinitely many) natural number $m$ such that $m$ leaves a remainder of 0 when divided by 49, a remainder of 1 when divided by 25, a remainder of 2 when divided by 9, and a remainder of 3 when divided by 4. (Then, as one can easily verify, $m>3$, so $m-3$ is also a natural number.)

Let $m$ be such a natural number. Then $m-3$ is divisible by $2^{2}$, $m-2$ by $3^{3}$, $m-1$ by $5^{2}$, and $m$ by $7^{2}$, so according to the previous lemma, each of these four consecutive natural numbers can be represented as the sum $x+y$ of positive integers and is simultaneously a divisor of the product $x \cdot y$ of these two summands. Therefore, such numbers exist.

Remark: The construction method can be extended in an analogous way to any number of consecutive natural numbers with this property.