Well, we see that with $a = 4, b = 2, c = 1$ we end up with $60 | n$. Now, those are all the positive integers $n$. Just see the numbers $a, b, c$ (it doesn't matter the order between them). Either one of those are $0$ in modulo $3$, or at least two of them are equal which implies that their difference is multiple of $3$. So, $3 | n$ For the modulos $4, 5$ is similar, if one of them is $0$ then we're done. If two of them are equal, their difference is $0$ and we're done. Then assuming that doesn't happen, in modulo $4$ we end up with {$a, b, c$} $=$ {$1, 2, 3$} and since $1 + 3$ is multiple of $4$, we're done. For modulo $5$, we conclude that {$a, b, c$} is exactly the set {$1, 2, 3, 4$} with one element deleted. So, at least one of the pairs {$1, 4$}, {$2, 3$} is contained in {$a, b, c$}. Its sum is multiple of $5$ and we're done. Therefore, we've proved that $60 | n$ is a necessary and sufficient condition. That's it!

Even though many people think of this problem as boring and trolling (you solve it with mods and not by manipulating expressions around divisibility), I personally prefer P1s to be such rather than monstrosities. Firstly, $n$ must be divisible by $3$, as if not, then $a^2 \equiv b^2 \equiv c^2 \equiv 1 \pmod 3$ and so $3 \mid a^2 - b^2 \mid n$, contradiction. Similarly, $n$ must be divisible by $4$ (otherwise at least two of $a$, $b$, $c$ are odd and have difference of squares divisible by $4$) and by $5$ (otherwise since $t^2 \equiv 1,4 \pmod 5$, at least two of $a^2$, $b^2$ and $c^2$ leave the same remainder modulo $5$, contradiction). Therefore $n$ must be divisible by $60$. On the other hand, for any $n$ divisible by $60$ we can take $a=4$, $b=2$, $c=1$ and note that $a^2 - b^2 = 12$, $b^2 - c^2 = 3$ and $a^2 - c^2 = 15$ divide $60$ and hence divide $n$.

This problem was proposed by me. I personally find it suitable for junior P1. It is easy to see that $n=60k$ for some positive integer $k$ works since we can take $a=4$, $b=2$ and $c=1$ because $4^2-2^2=12 \mid 60k$, $2^2-1^2 =3 \mid 60k$ and $4^2-1^2 =15 \mid 60k$. Now we will prove that all $n$ satisfying the conditions of the problem must be in this form. We make the following observations: $n$ is divisible by $3$. Suppose that $n$ is not divisible by $3$. Then $a^2 \equiv b^2 \equiv 1 \pmod 3$, hence $3 \mid a^2-b^2 \mid n$, which is a contradiction. Therefore $n$ is divisible by $3$. $n$ is divisible by $4$. Note that from the pigeonhole principle $2$ of the numbers $a,b,c$ have the same parity. WLOG assume that they are $a,b$. This means that $4 \mid (a+b)(a-b)=a^2-b^2 \mid n$, since $a+b$ and $a-b$ are both even. Therefore $n$ is divisible by $4$. $n$ is divisible by $5$. Suppose that $n$ is not divisible by $5$. It is a well-known fact that squares of positive integers are only $0,1$ or $4 \pmod 5$. However, none of the divisors is divisible by $5$. From the pigeonhole principle, $2$ of the numbers $a^2,b^2,c^2$ give the same residue when they are divided by $5$. WLOG assume that they are $a,b$. This means that $5 \mid a^2-b^2 \mid n$, hence $n$ is divisible by $5$. Since $n$ is divisible by $3,4$ and $5$, we can conclude that $60=3 \cdot 4 \cdot 5 \mid n$, hence $n=60k$ for some positive integer $k$, as desired.

@above nice problem. After noticing you just want there to be at most $3$ residues $\pmod{k}$ for $k|n$, and with one of them being $0$ which only happens if $x|k$ (if $k$ is prime), you're basically done. $x^2 \equiv -1,0,1 \pmod{k}$ happens for $k=3$ and $k=5$. The other case is when $k$ isn't prime but then you split into primes and you get $k=4$. So $60|n$ and take $a,b,c = (4,2,1)$, because you can apply pigeonhole principle on $(a^2,b^2,c^2)$ with two of them being equal $\pmod{k}$.

Claim 1: $ 3 \vert n$ quadratic residues on $3$ are $0$ and $1$. We have 3 variables $a^2$, $b^2$ and $c^2$, so in some pair residues will coincide, hence $3 \vert n$. Claim 2: $ 4 \vert n$ quadratic residues on $4$ are also $0$ and $1$. We can apply the same logic we did previously, hence $4 \vert n$. Claim 3: $ 5 \vert n$ quadratic residues on $5$ are $0$, $1$ and $4$. If WLOG $a^2 \equiv 0 \pmod{5}$, then $5 \vert a$ and $5 \vert n$. In other case, the residues will coincide on some pair and $5 \vert n$. Hence $60 \vert n$. For every $n$ multiple of $60$, we can take $c=1$, $b=2$ and $a=4$, which clearly works.

For $60 \mid n$, pick $a = 4$, $b = 2$, and $c = 1$. Note that $3 \mid n$, since if not, then $a^2 \equiv b^2 \equiv 1 \pmod{3}$, which is a contradiction. Also, note that if $n$ is odd, then $a^2 \equiv b^2 \pmod{4}$, leading to a contradiction. If $n$ is even, then if any two of $a$, $b$ are odd or any two of $a$, $b$ are even, then $4 \mid n$, which must happen by the pigeonhole principle, so $4 \mid n$. Now, if $5 \nmid n$, then $a$, $b$, and $c$ are congruent to $1$, $2$, $3$, and $4 \pmod{5}$, respectively. They are all distinct, but this leads to $5 \mid (a + b)$ or $5 \mid (b + c)$ or $5 \mid (a + c)$. So, $5 \mid n$ as well, and we are done.

Assassino9931 wrote: Determine all positive integers $n$ for which there exist positive divisors $a$, $b$, $c$ of $n$ such that $a>b>c$ and $a^2 - b^2$, $b^2 - c^2$, $a^2 - c^2$ are also divisors of $n$. The answer is all $n$ divisible by $60$. Clearly this works, since we can take $a=4,b=2,c=1$ to get the desired construction. Now we'll prove that they're the only solutions by showing that all such $n$ must be divisible by $60$. For any arbitary $x\in\mathbb{Z^+}$, we have $x\equiv 0,1\pmod{3}$ $x\equiv 0,1\pmod{4}$ $x\equiv 0,1,4\pmod{5}$ If $a^2,b^2,c^2$ have exactly the same value$\pmod{3}$, then all $a^2 - b^2$, $b^2 - c^2$, $a^2 - c^2$ are divisible by $3$, so $n$ is divisible by $3$. If not, we must have either one or two among $a^2,b^2,c^2$ are divisible by $3$, so $n$ is divisible by $3$. A similar argument shows that $n$ is divisible by $4$. If there are none of $a^2,b^2,c^2$ that are divisible by $5$, then by PHP, we must have at least two among $a^2,b^2,c^2$ having the same value$\pmod{5}$, so at least one among $a^2 - b^2$, $b^2 - c^2$, $a^2 - c^2$ are divisible by $5$. If there are at least one among $a,b,c$ that are divisible by $5$, then $n$ is divisible by $5$. Overall, $n$ is divisible by $3,4,5$, so $n$ is divisible by $60$, as desired. $\blacksquare$