Problem 2339

a.
Let us compute $D_n + D_{n+1}$ for $n = 35$, $76$, and $755$.

For $n = 35$:
$35$ is composite. Its largest proper divisor is $D_{35} = 35/5 = 7$ (since $35 = 5 \times 7$ and $7 < 35$).
$36$ is composite. Its largest proper divisor is $D_{36} = 36/2 = 18$ (since $36 = 2 \times 18$ and $18 < 36$).
So $D_{35} + D_{36} = 7 + 18 = 25 = 5^2$.

For $n = 76$:
$76$ is composite. Its largest proper divisor is $D_{76} = 76/2 = 38$.
$77$ is composite. Its largest proper divisor is $D_{77} = 77/7 = 11$.
So $D_{76} + D_{77} = 38 + 11 = 49 = 7^2$.

For $n = 755$:
$755$ is composite. Its largest proper divisor is $D_{755} = 755/5 = 151$.
$756$ is composite. Its largest proper divisor is $D_{756} = 756/2 = 378$.
So $D_{755} + D_{756} = 151 + 378 = 529 = 23^2$.

Therefore, $35$, $76$, and $755$ are squarish numbers.

b.
We want to show that there are infinitely many squarish numbers.

Let us look for $n$ such that $D_n + D_{n+1}$ is a perfect square.
Suppose $n$ is odd and $n+1$ is even.
For odd $n$, the largest proper divisor is $D_n = n/p$, where $p$ is the smallest prime dividing $n$.
For even $n+1$, $D_{n+1} = (n+1)/2$.

Let us try $n = p^2 - 1$, where $p$ is an odd prime.
Then $n = p^2 - 1$ is composite (since $p > 2$), and $n+1 = p^2$ is composite for $p > 2$.

$D_n = (p^2 - 1)/ (p-1) = p + 1$ (since $p^2 - 1 = (p-1)(p+1)$).
$D_{n+1} = D_{p^2} = p^2 / p = p$.
So $D_n + D_{n+1} = (p + 1) + p = 2p + 1$.

But $2p + 1$ is not always a perfect square. Let's try another approach.

Let $n$ be such that $n$ is divisible by $k$, and $n+1$ is divisible by $k+1$.
Suppose $n = k(k+1) - 1 = k^2 + k - 1$.
Then $n+1 = k^2 + k$.

$D_n = n / k = (k^2 + k - 1)/k = k + 1 - 1/k$ (not integer unless $k = 1$).

Alternatively, let us consider $n$ and $n+1$ such that $D_n = a$, $D_{n+1} = b$, and $a + b = m^2$.

Let us use the examples above:
$35 = 5 \times 7$, $D_{35} = 7$, $36 = 2 \times 18$, $D_{36} = 18$, $7 + 18 = 25$.
$76 = 2 \times 38$, $D_{76} = 38$, $77 = 7 \times 11$, $D_{77} = 11$, $38 + 11 = 49$.
$755 = 5 \times 151$, $D_{755} = 151$, $756 = 2 \times 378$, $D_{756} = 378$, $151 + 378 = 529$.

Notice that in each case, $n$ is divisible by $5$, $2$, $5$ respectively, and $n+1$ is divisible by $2$, $7$, $2$ respectively.

Let us generalize:
Let $n = p \times q$, $D_n = q$ (assuming $p < q$), $n+1 = 2 \times r$, $D_{n+1} = r$.
Then $q + r = m^2$.

Alternatively, for $n$ odd, $n+1$ even, $D_{n+1} = (n+1)/2$.
Suppose $n = k \times m$, $D_n = m$.
Let $n+1 = 2 \times t$, $D_{n+1} = t$.
So $m + t = s^2$.

Let us try $n = 2k - 1$, $n+1 = 2k$.
$D_n$ depends on the factorization of $2k - 1$.
$D_{n+1} = k$.
If $2k - 1$ is composite and its largest proper divisor is $m$, then $m + k$ is a perfect square for infinitely many $k$.

Alternatively, for $n$ even, $n = 2k$, $D_n = k$, $n+1$ odd, $D_{n+1}$ depends on its factorization.

But from the examples, we see that for $n = 5 \times 7 = 35$, $n+1 = 36 = 2 \times 18$, $D_{35} = 7$, $D_{36} = 18$, $7 + 18 = 25$.
Similarly, for $n = 5 \times p$, $n+1 = 2 \times q$, $D_n = p$, $D_{n+1} = q$, $p + q = m^2$.

Therefore, for any $m$, let $p = m^2 - q$, $q$ arbitrary, $n = 5 \times p$, $n+1 = 2 \times q$, and $p + q = m^2$.

Thus, there are infinitely many squarish numbers.

Alternatively, since the set of perfect squares is infinite, and for each perfect square $m^2$, we can find $n$ such that $D_n + D_{n+1} = m^2$, there are infinitely many squarish numbers.