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Max Alekseyev
Max Alekseyev
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First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$). As divisors of pairwise coprime numbers, $a+b$, $a+c$, $b+c$ are also pairwise coprime.

Now, since $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid c^2$, each of $a+b$, $a+c$, $b+c$ divides $$D = (a+b)^2 + (a+c)^2 + (b+c)^2 - a^2 - b^2 - c^2.$$ Then their product $(a+b)(a+c)(b+c)$ must also divide $D$.

Without loss of generality, assume that $a\leq b\leq c$ and so $a+b\leq a+c\leq b+c$. Then $(a+b)(a+c)(b+c) \leq D < 3(b+c)^2$ and hence $$(a+b)c < (a+b)(a+c) < 3(b+c) \leq 6c,$$ implying that $a+b < 6$.

So, there is a finite number of cases to consider. A quick computation revealedIt is easy to check that none of them gives a solution. Namely, for any fixed $a,b$, possible $c$ must belong to the finite set: $$\{ d-b\ :\ d\mid a^2\} \cap \{ d-a\ :\ d\mid b^2\}.$$

First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$). As divisors of pairwise coprime numbers, $a+b$, $a+c$, $b+c$ are also pairwise coprime.

Now, since $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid c^2$, each of $a+b$, $a+c$, $b+c$ divides $$D = (a+b)^2 + (a+c)^2 + (b+c)^2 - a^2 - b^2 - c^2.$$ Then their product $(a+b)(a+c)(b+c)$ must also divide $D$.

Without loss of generality, assume that $a\leq b\leq c$ and so $a+b\leq a+c\leq b+c$. Then $(a+b)(a+c)(b+c) \leq D < 3(b+c)^2$ and hence $$(a+b)c < (a+b)(a+c) < 3(b+c) \leq 6c,$$ implying that $a+b < 6$.

So, there is a finite number of cases to consider. A quick computation revealed that none of them gives a solution.

First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$). As divisors of pairwise coprime numbers, $a+b$, $a+c$, $b+c$ are also pairwise coprime.

Now, since $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid c^2$, each of $a+b$, $a+c$, $b+c$ divides $$D = (a+b)^2 + (a+c)^2 + (b+c)^2 - a^2 - b^2 - c^2.$$ Then their product $(a+b)(a+c)(b+c)$ must also divide $D$.

Without loss of generality, assume that $a\leq b\leq c$ and so $a+b\leq a+c\leq b+c$. Then $(a+b)(a+c)(b+c) \leq D < 3(b+c)^2$ and hence $$(a+b)c < (a+b)(a+c) < 3(b+c) \leq 6c,$$ implying that $a+b < 6$.

So, there is a finite number of cases to consider. It is easy to check that none of them gives a solution. Namely, for any fixed $a,b$, possible $c$ must belong to the finite set: $$\{ d-b\ :\ d\mid a^2\} \cap \{ d-a\ :\ d\mid b^2\}.$$

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Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152

First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$). SimilarlyAs divisors of pairwise coprime numbers, we have that $a+b$, $a+c$, $b+c$ are also pairwise coprime.

Now, ifsince $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid c^2$, theneach of $(a+b)(a+c)(b+c)$ divides$a+b$, $a+c$, $b+c$ divides $$D = (a+b)^2 + (a+c)^2 + (b+c)^2 - a^2 - b^2 - c^2.$$ Then their product $(a+b)(a+c)(b+c)$ must also divide $D$.

Without loss of generality, assume that $a\leq b\leq c$ and so $a+b\leq a+c\leq b+c$. Then $(a+b)(a+c)(b+c) \leq D < 3(b+c)^2$ and hence $$(a+b)c < (a+b)(a+c) < 3(b+c) \leq 6c,$$ implying that $a+b < 6$.

So, there is a finite number of cases to consider. A quick computation revealed that none of them gives a solution.

First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$). Similarly, we have that $a+b$, $a+c$, $b+c$ are pairwise coprime.

Now, if $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid c^2$, then $(a+b)(a+c)(b+c)$ divides $$D = (a+b)^2 + (a+c)^2 + (b+c)^2 - a^2 - b^2 - c^2.$$

Without loss of generality, assume that $a\leq b\leq c$ and so $a+b\leq a+c\leq b+c$. Then $(a+b)(a+c)(b+c) \leq D < 3(b+c)^2$ and hence $$(a+b)c < (a+b)(a+c) < 3(b+c) \leq 6c,$$ implying that $a+b < 6$.

So, there is a finite number of cases to consider. A quick computation revealed that none of them gives a solution.

First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$). As divisors of pairwise coprime numbers, $a+b$, $a+c$, $b+c$ are also pairwise coprime.

Now, since $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid c^2$, each of $a+b$, $a+c$, $b+c$ divides $$D = (a+b)^2 + (a+c)^2 + (b+c)^2 - a^2 - b^2 - c^2.$$ Then their product $(a+b)(a+c)(b+c)$ must also divide $D$.

Without loss of generality, assume that $a\leq b\leq c$ and so $a+b\leq a+c\leq b+c$. Then $(a+b)(a+c)(b+c) \leq D < 3(b+c)^2$ and hence $$(a+b)c < (a+b)(a+c) < 3(b+c) \leq 6c,$$ implying that $a+b < 6$.

So, there is a finite number of cases to consider. A quick computation revealed that none of them gives a solution.

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Max Alekseyev
Max Alekseyev
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First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$). Similarly, we have that $a+b$, $a+c$, $b+c$ are pairwise coprime.

Now, if $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid c^2$, then $(a+b)(a+c)(b+c)$ divides $$D = (a+b)^2 + (a+c)^2 + (b+c)^2 - a^2 - b^2 - c^2.$$

SupposeWithout loss of generality, assume that $a\leq b\leq c$ and so $a+b\leq a+c\leq b+c$. Then $(a+b)(a+c)(b+c) \leq D < 3(b+c)^2$ and hence $$(a+b)c < (a+b)(a+c) < 3(b+c) \leq 6c,$$ implying that $a+b < 6$.

So, there is a finite number of cases to consider. A quick computation revealed that none of them gives a solution.

First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$). Similarly, we have that $a+b$, $a+c$, $b+c$ are pairwise coprime.

Now, if $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid c^2$, then $(a+b)(a+c)(b+c)$ divides $$D = (a+b)^2 + (a+c)^2 + (b+c)^2 - a^2 - b^2 - c^2.$$

Suppose that $a\leq b\leq c$ and so $a+b\leq a+c\leq b+c$. Then $(a+b)(a+c)(b+c) \leq D < 3(b+c)^2$ and hence $$(a+b)c < (a+b)(a+c) < 3(b+c) \leq 6c,$$ implying that $a+b < 6$.

So, there is a finite number of cases to consider. A quick computation revealed that none of them gives a solution.

First, notice that such $a$, $b$, $c$ must be pairwise coprime (e.g., if prime $p\mid \gcd(a,b)$, then $(a+b)\mid c^2$ implies $p\mid c$, a contradiction to $\gcd(a,b,c)=1$). Similarly, we have that $a+b$, $a+c$, $b+c$ are pairwise coprime.

Now, if $(a+b)\mid c^2$, $(a+c)\mid b^2$, $(b+c)\mid c^2$, then $(a+b)(a+c)(b+c)$ divides $$D = (a+b)^2 + (a+c)^2 + (b+c)^2 - a^2 - b^2 - c^2.$$

Without loss of generality, assume that $a\leq b\leq c$ and so $a+b\leq a+c\leq b+c$. Then $(a+b)(a+c)(b+c) \leq D < 3(b+c)^2$ and hence $$(a+b)c < (a+b)(a+c) < 3(b+c) \leq 6c,$$ implying that $a+b < 6$.

So, there is a finite number of cases to consider. A quick computation revealed that none of them gives a solution.

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Max Alekseyev
Max Alekseyev
  • 34.3k
  • 5
  • 74
  • 152