If $a+b+c$ divides the product $abc$, then is $(a,b,c)$ a Pythagorean Triple?

Firstly, I will define what Pythagorean Triples are for those who do not know.

Definition:

A Pythagorean Triple is a group of three integers $a$, $b$ and $c$ such that $a^2+b^2=c^2$, since the Pythagorean Theorem asserts that for any $90^\circ$ (right-angle) triangle $ABC$ with sides $a$, $b$ and $c$, one will always have the equation, $a^2+b^2=c^2$. I was looking at Pythagorean Triples and noticed another property apart from how $a^2+b^2=c^2$. Here are the first $30$ Pythagorean Triples $(a,b,c)$ ordered from smallest to greatest value, i.e. $$(a,b,c)\qquad\text{ s.t. }\qquad a<b<c.\tag*{\big(\text{s.t. = such that}\big)}$$ I noticed that $a^2=(c+b)(c-b)$, but that is trivial since \begin{align}a^2&=(c+b)(c-b)\tag{given} \\ &=c^2-b^2 \\ \Leftrightarrow\,\,\,\, a^2+b^2&=c^2.\end{align}

However, I also noticed that by having "$u\mid v$" be read as "$u$ divides $v$", it appears that $$a+b+c\mid abc.$$ For example, $(a,b,c)=(3,4,5)$ is a classic Pythagorean Triple; $3^2+4^2=5^2$.

Also, \begin{align}3+4+5&=12 \\ \& \quad3\times 4\times 5 &= 60. \\ \\ 12 &\,\mid 60 \\ \Leftrightarrow \,\,\,\,3+4+5&\,\mid 3\times 4\times 5.\end{align} This, I cannot prove to be true $-$ but I tested with all the $30$ Pythagorean Triples above, and I have come across no counter-example. Is there a proof? I do not know where to begin myself.

Conjecture:

Given three positive integers $a$, $b$ and $c$, if $a < b<c$ and $a^2+b^2=c^2$, then $$a+b+c\mid abc.$$

Thank you in advance.

Edit:

My conjecture was originally the other way round; i.e. if $a+b+c\mid abc$ then $a^2+b^2=c^2$. But $6$ is a counter-example, namely because it is a Perfect Number.

• Take $(a,b,c)=(1,2,3)$ in the conjecture. Then $a+b+c=6$ divides $abc=6$, but $1^2+2^2\neq 3^2$. – Dietrich Burde Jun 20 '18 at 11:18
• Wow that is a counter-example! Looks like I have to restate my conjecture :) – Mr Pie Jun 20 '18 at 11:21
• @user477343 perhaps you want the converse; in your example you've taken a Pythagorean triple and verified that the desired property holds. – ÍgjøgnumMeg Jun 20 '18 at 11:22
• Newton's polynoms could design different things : – Pagode Jun 20 '18 at 11:29
• – Pagode Jun 20 '18 at 11:32

You actually want it the other way around: if $a^2+b^2=c^2$ then $a+b+c|abc$. That you can prove very quickly from the general form of primitive Pythagorean triples $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$.

• \begin{align}a+b+c&=(m^2-n^2)+(2mn)+(m^2+n^2)\\ &=(m^2+m^2)+2mn+(n^2-n^2) \\ &=2m^2+2mn \\ &=2m(m+n).\end{align} And now $abc=2mn(m^4-n^4)$ so I must show that $m+n\mid n(m^4-n^4)$ which means $n\mid m$. But $m$ and $n$ are arbitrary. Am I doing something wrong? – Mr Pie Jun 20 '18 at 11:30
• $x^4-y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x ^2+y^2)$ – Stefan Jun 20 '18 at 11:38
• $a+b+c=2m^2+2mn=2m(m+n)$, while $abc={\bf 2m}n(m^2+n^2)(m-n){\bf (m+n)}$. – Berci Jun 20 '18 at 11:40
• @Stefan thank you very much. I was able to put to-and-to together :) – Mr Pie Jun 20 '18 at 11:42
• Congratulations, Michael! You have a tick! $$\color{green}{\checkmark} \ \ (+1)$$ – Mr Pie Jun 20 '18 at 11:43

Alternatively: $$\frac{abc}{a+b+c}=\frac{abc(a+b-c)}{(a+b+c)(a+b-c)}=\frac{abc(a+b-c)}{2ab}=\frac{c(a+b-c)}{2},$$ which is a positive integer for both cases: $a,b,c$ are all even; $a,c$ are odd and $b$ is even.

• I am glad you used a conjugate method. I only use those to rationalise denominators. I did not know you could use that technique in this case. $(+1)$. There is, however, also another case, where $c$ is even and $a+b>c$, such that $a$ and $b$ are both odd or even. – Mr Pie Jun 20 '18 at 12:13
• It follows that $a+b+c\mid ab$. – g.kov Jun 20 '18 at 17:26
• Also, the integer factor $\tfrac12(a+b-c)=r$ is the radius of inscribed circle. – g.kov Jun 20 '18 at 17:35
• @g.kov I knew the first part... but not the second. Is the inscribed circle the circle drawn inside the triangle such that the circumference is just touching the legs and hypotenuse? I assume I understand what "inscribed" means, hahah :) – Mr Pie Jun 20 '18 at 22:13