Divisibility of a triple

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goodar2006

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goodar2006

#1 h Apr 19, 2013, 11:43 AM • 9 Y

Y by mlm95, shekast-istadegi, mahanmath, Fil757, Davi-8191, doxuanlong15052000, Adventure10, Mango247, Efesc128e968

Do there exist natural numbers $a, b$ and $c$ such that $a^2+b^2+c^2$ is divisible by $2013(ab+bc+ca)$ ?

Proposed by Mahan Malihi

This post has been edited 1 time. Last edited by goodar2006, Apr 21, 2013, 1:54 AM

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apluscactus

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apluscactus

#2 h Apr 19, 2013, 4:40 PM • 2 Y

Y by rightways, Adventure10

One can prove that $\frac{a^{2}+b^{2}+c^{2}}{ab+bc+ca}$ is never divisible by 3.

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v_Enhance

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v_Enhance

#3 h Apr 19, 2013, 6:32 PM • 15 Y

Y by narutomath96, ssilwa, AnonymousBunny, kun1417, nmd27082001, Anar24, darkeagle, Polynom_Efendi, Illuzion, IAmTheHazard, Kobayashi, Aopamy, honey_lemon, Adventure10, Mango247

Assume $a,b,c$ do not have some common divisor and write $(2013k+2)(a^2+b^2+c^2) = 2013k \cdot (a+b+c)^2.$ Since $2013k + 2 \equiv 2 \pmod{3}$ , there is a prime $p \equiv 2 \pmod{3}$ with $v_p(2013k+2)$ odd (in particular, $p \mid 2013k+2$ ).

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Claim: $p \neq 2$ .
Let us assume on the contrary that $v_2(2013k+2)$ is odd (in particular, $k$ is even). Remark that since $a,b,c$ are not all even, $v_2(a^2+b^2+c^2) \le 1$ . Furthermore, $v_2(a+b+c) = 0 \iff v_2(a^2+b^2+c^2)=0.$ Now we consider two cases.

Case 1: $v_2(k) \ge 2$ . Then $v_2(2013k+2)=1$ , $v_2(a^2+b^2+c^2) \le 1$ . Therefore $v_2(k) + 2v_2(a+b+c) \ge 2 \ge v_2(2013k+2) + v_2(a^2+b^2+c^2)$ but equality cannot occur since the relations $v_2(a+b+c) = 0 \quad \text{and} \quad v_2(a^2+b^2+c^2) = 1$ cannot hold simultaneously.
Case 2: $v_2(k) = 1$ . Then $v_2(2013k+2) > 1$ and is odd. Now $v_2(2013k+2) + v_2(a^2+b^2+c^2) = 1 + 2v_2(a+b+c).$ Considering mod 2, we see $v_2(a^2+b^2+c^2)$ must be even, so it is zero; consequently $v_2(a+b+c) = 0$ as well and we obtain $1 < v_2(2013k+2) = 1$ .

===============================

Now for the interesting part. Remark $p \mid a+b+c$ and $p \mid a^2+b^2+c^2$ . Without loss of generality $b \not\equiv 0 \pmod{p}$ , so that $a^2+b^2+(a+b)^2 \equiv 0 \pmod{p} \implies a^2+ab+b^2 \equiv 0 \pmod{p}.$ Then if $x = ab^{-1}$ , we get that $x^2+x+1 \equiv 0 \pmod{p}$ . But the LHS is the third cyclotomic polynomial, so either $p=3$ or $3 \mid p-1$ , but neither is the case.

Therefore, such a triple $(a,b,c)$ does not exist.

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dinoboy

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dinoboy

#4 h Apr 19, 2013, 6:58 PM • 4 Y

Y by narutomath96, Binomial-theorem, Adventure10, Mango247

Let $a^2 + b^2 + c^2 = k(ab + ac + bc)$ . Looking at this as a quadratic in $c$ , we have $c^2 - c(ka + kb) + (a^2 + b^2 - kab) = 0$ . Thus we can vieta-jump from a solution $(a,b,c)$ to $\left (a,b, \frac{a^2 + b^2 - kab}{c} \right )$ . Thus we can WLOG $c \le \sqrt{a^2 + b^2}$ and $a \le b \le c$ (just take the minimum solution) unless $c = ka + kb$ . We'll deal with that case in a bit.
Now note that:

$\begin{eqnarray*} \frac{a^2 + b^2 + c^2}{ab + ac + bc} & \le & \frac{2a^2 + 2b^2}{ab + ac + bc} \\ & \le & \frac{2a^2 + 2b^2}{ab + a^2 + b^2} \\ & \le & 2 \end{eqnarray*}$

So obviously in this case $a^2 + b^2 + c^2$ is not divisible by $2013(ab + ac + bc)$ . Now suppose $c = ka + kb$ . We have:

$\begin{eqnarray*} \frac{a^2 + b^2 + c^2}{ab + ac + bc} & = & \frac{a^2 + b^2 + k^2(a+b)^2}{ab + k(a+b)^2} \\ & = & k + \frac{a^2 + b^2 - kab}{ab + k(a+b)^2} \\ & = & k \end{eqnarray*}$

due to $a^2 + b^2 - kab = 0$ as well. Thus it suffices to show $a^2 + b^2 = kab$ has no solutions for $k = 2013$ . However, to do this just vieta jump again to get the minimum solution has $a=b$ and thus $k=2$ . Therefore in fact if $(ab+ac+bc)|(a^2 + b^2 + c^2)$ , then $a^2 + b^2 + c^2$ is either double $ab + ac + bc$ or equal to it which shows the result.

EDIT : oops I forgot that $c$ can be greater than $ka + kb$ darn... disregard this solution.
Apparently $\frac{a^2 + b^2+c^2}{ab + ac + bc} =1,2,5,10,14$ based on a program. I think some trickier vieta jumping can make this solution work.

This post has been edited 1 time. Last edited by dinoboy, Apr 19, 2013, 7:25 PM

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Goutham

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Goutham

#5 h Apr 19, 2013, 7:20 PM • 2 Y

Y by Adventure10, Mango247

@dinoboy, I did almost exactly as you did but I got stuck at this point.

dinoboy wrote:

Thus we can vieta-jump from a solution $(a,b,c)$ to $\left (a,b, \frac{a^2 + b^2 - kab}{c} \right )$ .

What happens if $a^2+b^2 < kab$ ? It happens for $(a, b, c) = (3, 5, 41)$ .

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v_Enhance

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v_Enhance

#6 h Apr 19, 2013, 7:23 PM • 2 Y

Y by Adventure10, Mango247

Indeed, dinoboy's solution claims that $ab + bc + ca \mid a^2 + b^2 + c^2 \implies k=2$ , but this is false for many triples.

evan@ArchAir ~/Downloads
$ python
Python 2.7.4 (default, Apr  6 2013, 19:20:36) 
[GCC 4.8.0] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import itertools
>>> for a,b,c in itertools.product(range(1,100), repeat=3):
...     m = a*a + b*b + c*c
...     n = a*b + b*c + c*a
...     if m % n != 0: continue
...     if m/n > 2:
...             print a,b,c
...

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dinoboy

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dinoboy

#7 h Apr 19, 2013, 7:31 PM • 2 Y

Y by Adventure10, Mango247

Ok, so I messed up. darn. However, one thing I find that is it appears if $k > 2$ then usually $c = ka + kb + 1$ in the minimum solution (. Can anybody prove this? As this would lead to a solution.

Or better, if somebody could prove apluscactus's claim that would be nice.

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apluscactus

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apluscactus

#8 h Apr 19, 2013, 7:41 PM • 2 Y

Y by Adventure10, Mango247

The solution proposed by V_ENhance prouves my claim : just replace 2013 by any multiple of 3. This was my solution to.

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dinoboy

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dinoboy

#9 h Apr 19, 2013, 7:43 PM • 2 Y

Y by Adventure10, Mango247

But the problem is that his solution is incomplete...

Also, one thing to note is that $v_3 \left ( \frac{a^2 + b^2 + c^2}{ab + ac + bc} \right ) = 1$ is possible if you remove the restriction it must be an integer (just see $a=1, b = 8, c = 88$ ) so arguments looking purely at the $3$ -adic evalution fail.

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mathocean97

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mathocean97

#10 h Apr 19, 2013, 9:02 PM • 2 Y

Y by narutomath96, Adventure10

This is similar to v_enhance's solutions.

First note that if $p$ divides 2 of $a,b,c$ , the $p$ divides the last one.
So we let $a^2 + b^2 + c^2 = k(ab+bc+ac)$ , where $k$ is a multiple of 3.
Assume $gcd(a, b, c) = 1$ .
This rearranges as $a^2 - a\left(k(b+c)\right) + b^2-kbc+c^2 = 0$ .
So the discriminant of the quadratic is

$(k^2-4)b^2 + (2k^2+4k)bc + (k^2-4)c^2 = (k+2)\left((k-2)b^2 + 2kbc + (k-2)c^2\right)$ .
Since $k+2 \equiv 2 \pmod{3}$ , there exists a prime $p$ dividing $k+2$ with odd multiplicity.

Distinguish 2 cases:
Case 1: $p$ is odd.

Then $p \mid \left((k-2)b^2 + 2kbc + (k-2)c^2\right)$ , and since $p \mid k+2$ ,
$(k-2)b^2 + 2kbc + (k-2)c^2 \equiv -4b^2-4bc-4c^2 \pmod{p}$

So $p \mid b^2+bc+c^2 \mid b^3-c^3$ . Since $p \equiv 2 \pmod{3}$ , all residues $x^3 \pmod{p}$ are distinct, so $p \mid b-c$ .
If $p \mid b$ and $p \mid c$ , then we can divide $a, b, c$ by $p$ and get a smaller triple.
Otherwise, $\upsilon_p(b^2+bc+c^2) = \upsilon_p(b^3-c^3) - \upsilon_p(b-c) = \upsilon_p(b-c) + \upsilon_p(3) -\upsilon_p(b-c) = 0$ , by the Lifting Exponent Lemma. So this case is impossible.

Case 2: $p = 2$ .
Since we know that the multiplicity of 2 in $(k+2)$ is odd, we can write $k = j\cdot 2^{2m+1} - 2$ , where $j$ is odd and $m \ge 0$ .

Assume $m = 0$ . Then clearly $4 \vert k$ . But going back to the beginning, we get that $4 \mid a^2+b^2+c^2$ , so $a,b,c$ are all even, contradiction.
Now let $m \ge 1$ .
Then note that
$((k-2)b^2+2kbc+(k-2)c^2)$
$= (j\cdot 2^{2m+1} - 4)b^2+(j\cdot 2^{2m+2} - 4)bc+(j\cdot 2^{2m+1} - 4)c^2$
$= 4((j\cdot 2^{2m-1} - 1)b^2 + (j\cdot 2^{2m} - 1)bc + (j\cdot 2^{2m-1} - 1)c^2)$

So we still need $2 \mid \left((j\cdot 2^{2m-1} - 1)b^2 + (j\cdot 2^{2m} - 1)bc + (j\cdot 2^{2m-1}-1)c^2\right)$ . But the coefficients of $b^2, bc, c^2$ are all odd, so it is easy to check that it is even iff $b, c$ are both even. Then $a, b, c$ are all even again, contradiction.

Remark: In fact this shows that $3 \not\vert k$ and $\upsilon_2(k+2) \not\equiv 1 \pmod{2}$ .

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siddigss

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siddigss

#11 h Apr 19, 2013, 9:50 PM • 1 Y

Y by Adventure10

$11|a^2+b^2+c^2\Rightarrow 11|\gcd(a,b,c)$ ???!!!

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v_Enhance

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v_Enhance

#12 h Apr 19, 2013, 10:50 PM • 2 Y

Y by Adventure10, Mango247

Finished my solution. During chemistry I unfortunately did not realize that $v_2$ was in fact sufficient. By the way, thanks to whichever moderator fixed a typo for me

siddigss wrote:

$11|a^2+b^2+c^2\Rightarrow 11|\gcd(a,b,c)$ ???!!!

Sorry, what is this referring to? Anyway, it is true in the context of this problem (namely, $11 \mid a+b+c$ and $11 \mid a^2+b^2+c^2$ implies $a,b,c \equiv 0 \pmod{11}$ . See the last paragraph of my solution.)

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siddigss

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siddigss

#13 h Apr 20, 2013, 2:44 AM • 2 Y

Y by Adventure10, Mango247

v_Enhance wrote:

Sorry, what is this referring to? Anyway, it is true in the context of this problem (namely, $11 \mid a+b+c$ and $11 \mid a^2+b^2+c^2$ implies $a,b,c \equiv 0 \pmod{11}$ . See the last paragraph of my solution.)

I exactly meant what I've written $11|a^2+b^2+c^2$ implies $a,b,c\equiv0(\mod 11)$ even if $11\not|a+b+c$ -using it the problem answer is NO , because there is no 3 coprimes such that the sum of their squares is divisible by 11 ( $11|2013$ ) -( but we didn't use $ab+bc+ca|a^2+b^2+c^2$ !! )

thanks v_Enhance

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v_Enhance

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v_Enhance

#14 h Apr 20, 2013, 3:00 AM • 3 Y

Y by siddigss, Adventure10, Mango247

siddigss wrote:

I exactly meant what I've written $11|a^2+b^2+c^2$ implies $a,b,c\equiv0(\mod 11)$ even if $11\not|a+b+c$

Oh, sorry, I thought you were asking a question about the solution above yours. But I don't think this claim is true: $11 \mid 3^2 + 1^2 + 1^2$ , for instance.

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siddigss

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#15 h Apr 20, 2013, 3:08 AM • 2 Y

Y by Adventure10, Mango247

Yeah yeah sorry

unforgivable mistake .. Thanks

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rama1728

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rama1728

#41 h Sep 16, 2021, 6:56 AM

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goodar2006 wrote:

Do there exist natural numbers $a, b$ and $c$ such that $a^2+b^2+c^2$ is divisible by $2013(ab+bc+ca)$ ?

Proposed by Mahan Malihi

We show that the quantity $\frac{a^2+b^2+c^2}{3(ab+bc+ca)}$ is never an integer. Then we are done.

Assume that $a^2+b^2+c^2=3n(ab+bc+ca)$ then $(a+b+c)^2=(3n+2)(ab+bc+ca)$ Now choose a prime $p$ congruent to $2\pmod3$ so that $p\mid3n+2$ . Then, we have that for some $i$ , $p^i\mid a+b+c$ and $p\mid ab+bc+ca$ (simply choose an odd power of $p$ dividing $3n+2$ ). So, $c\equiv-a-b\pmod p$ and substituting in the other relation gives $p\mid a^2+ab+b^2$ or $p\mid (2a+b)^2+3b^2$ so by Quadratic Residue, $\left(-3\over p\right)=1$ which is a contradiction since $p\equiv2\pmod3$ .

This post has been edited 1 time. Last edited by rama1728, Sep 16, 2021, 6:57 AM
Reason: .

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guptaamitu1

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guptaamitu1

#42 h Feb 9, 2022, 11:18 AM

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The answer is NO. Assume contrary. WLOG $\gcd(a,b,c) =1$ and $a \ge b \ge c$ . Let $x=a-b,y=b-c$ . Write
$\begin{align*} a^2 + b^2 + c^2 &= 2013k(ab + bc + ca) \\ \implies a^2 + b^2 + c^2 - (ab + bc + ca) &= (2013k-1)(ab + bc + ca) \\ \implies (2013k-1)(ab+bc+ca) &= \frac{1}{2} \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right) \\ &= \frac{1}{2} \left( x^2 + y^2 + (x+y)^2 \right) = x^2 + y^2 + xy \\ \implies x^2 + y^2 + xy &= (2013k-1)(ab + bc + ca) \qquad \qquad (1) \end{align*}$ Pick a prime $p \equiv 2 \pmod{3}$ such that $v_p(2013k-1)$ is even (such a $p$ exists as $2013k-1 \equiv 2 \pmod{3}$ ). Write $x = dm,y=dn$ with $\gcd(m,n) = 1$ . Note $p$ cannot divide both $m,n$ .

Claim: $p \mid m^2 + n^2 + mn$ .
Proof: If not, then $p \mid x,y$ and also $p \mid ab + bc + ca$ . From $p \mid x,y$ we get $a \equiv b \equiv c \pmod{p}$ . But as $p \mid ab + bc + ca$ , so this forces $p \mid a,b,c$ , which contradicts $\gcd(a,b,c) = 1$ . $\square$

Now $4(m^2 + n^2 + mn) = (2m+n)^2 + 3n^2$ . Hence,
$(2m+n)^2 + 3n^2 \equiv 0 \pmod{p}$ As $p \nmid m,n$ , so $-3$ is a QR mod $p$ . But as $p \equiv 2 \pmod{3}$ , so that's a contradiction. $\blacksquare$

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eduD_looC

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eduD_looC

#43 h Mar 18, 2023, 5:40 PM

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Can someone explain why if $p \equiv 2 \pmod 3$ , then $-3$ cannot be a is a QR mod $p$ ? I'm looking at $p=2$ , and $-3 \equiv 1$ seems to be a QR.

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eduD_looC

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eduD_looC

#44 h Mar 19, 2023, 9:01 PM

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bump bump

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eduD_looC

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eduD_looC

#45 h Mar 22, 2023, 12:54 AM

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bump bump

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v_Enhance

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v_Enhance

#46 h Mar 22, 2023, 1:47 AM • 1 Y

Y by HamstPan38825

eduD_looC wrote:

Can someone explain why if $p \equiv 2 \pmod 3$ , then $-3$ cannot be a is a QR mod $p$ ? I'm looking at $p=2$ , and $-3 \equiv 1$ seems to be a QR.

That statement only holds for odd primes $p$ . It follows by quadratic reciprocity theorem.

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eduD_looC

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eduD_looC

#47 h Mar 23, 2023, 1:43 AM

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v_Enhance wrote:

eduD_looC wrote:

Can someone explain why if $p \equiv 2 \pmod 3$ , then $-3$ cannot be a is a QR mod $p$ ? I'm looking at $p=2$ , and $-3 \equiv 1$ seems to be a QR.

That statement only holds for odd primes $p$ . It follows by quadratic reciprocity theorem.

Yes, but then how is the $p=2$ case dealt with? The last two solutions seem to disregard that case, so I was wondering if there were conditions I was missing out that would eliminate this extra case.

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DottedCaculator

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DottedCaculator

#48 h Mar 23, 2023, 2:08 AM

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$p=2$ can be solved directly by noting $2\mid x^2+xy+y^2$ implies $2\mid x,y$

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HamstPan38825

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#49 h Apr 23, 2023, 3:11 PM

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We prove the following general problem: if $a^2+b^2+c^2=k(ab+bc+ca)$ for some integer $k$ , then $3 \nmid k$ .

Assume otherwise. WLOG we may assume $\gcd(a, b, c) = 1$ . Then, $(a+b+c)^2=(3n+2)(ab+bc+ca),$ so there exists a $3k+2$ form prime $p \mid 3n+2$ . Now, note that $\nu_p((a+b+c)^2)$ is even and $\nu_p(3n+2)$ is odd, thus $p \mid ab+bc+ca$ . But now $a^2+ab+b^2 \equiv ab+bc+ca \equiv 0 \pmod p,$ and thus we must have $p \mid a, b, c$ , which contradicts $\gcd(a, b, c) = 1$ .

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ihategeo_1969

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ihategeo_1969

#50 h Oct 15, 2023, 12:21 AM

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We claim there are no solutions to this equation. Assume the contrary.

See that we can rewrite the problem statement as ${(a+b+c)}^2=(2013n+2)(ab+bc+ca)$ where $n \in \mathbb{Z}$ and WLOG assume $\gcd(a,b,c)=1$ .
Claim We cannot have $2 \mid a+b+c$ and $2 \mid ab+bc+ca$ .
Subproof. See that since of our assumption, we must have two of $a$ , $b$ and $c$ to be odd and last one to be even but this means $ab+bc+ca$ must be odd. $\blacksquare$

See that $2013n+2 \equiv 2 \pmod 3$ , and hence there exists a prime $p \equiv 2 \pmod 3$ such that $v_p(2013n+2)$ is odd which means $p \mid a+b+c$ but $v_p \left({(a+b+c)}^2 \right)$ is even and hence we get $p \mid ab+bc+ca$ (see that $p \geq 5$ ).

See that if $p$ divides any of $a$ , $b$ or $c$ then it divides all of them, a contradiction. And now considering $c \equiv -a-b \pmod p$ and let $x \equiv \frac{a}b \pmod p$ , we get $p \mid x^2+x+1 \Longleftrightarrow {(2x+1)}^2 \equiv -3 \pmod p \Longleftrightarrow \left( \frac{-3}p \right) \Longleftrightarrow p \equiv 1 \pmod 3$ which is a contradiction.

This post has been edited 1 time. Last edited by ihategeo_1969, Apr 30, 2024, 7:24 PM
Reason: 13434

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math_comb01

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math_comb01

#51 h Dec 10, 2023, 1:25 PM

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Nice Problem.
Assume $(a,b,c)=1$
Let $a^2+b^2+c^2 = 3k(ab+bc+ca)$ as $3|2013$
then $(a+b+c)^2 = (3k+2)(ab+bc+ca)$ , now there must exist a prime $p$ dividing $3k+2$ such that $v_p(3k+2)$ is odd.
$p|(a+b+c)^2$ $\rightarrow$ $p|a+b+c$ and as $v_p(3k+2)$ is odd so $p|ab+bc+ca$
$\rightarrow p|(a+b)(-a-b)+ab$
and hence $p|a^2+b^2+ab$ if $p$ does not divide $a,b$ then
$(2a+b)^2 \equiv -3b^2 (\mod p)$
$(2a \cdot b^{-1}+1)^2 \equiv -3 (\mod p)$
$1=\left(\frac{-3}{p}\right)= (-1)^{\frac{p-1}{2}} \cdot \left(\frac{3}{p}\right)=(-1)^{\frac{p-1}{2}} \cdot \left(\frac{p}{3}\right) \cdot (-1)^{\frac{p-1}{2}}=-1$ This yields a contradiction as desired.

This post has been edited 1 time. Last edited by math_comb01, Dec 10, 2023, 5:03 PM

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Markas

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#52 h Jul 13, 2024, 2:28 PM

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Assume gcd(a,b,c) = 1. Now let $n = \frac{a^2 + b^2 + c^2}{2013(ab + bc + ac)}$ , which is equivalent to $(a + b + c)^2 = (2013n + 2)(ab + bc + ca)$ . Since $2013n + 2 \equiv 2 \pmod 3$ , then $p \mid 2013n + 2$ , where $p \equiv 2 \pmod 3$ and $\nu_p(2013n + 2) = odd$ . Since $(a + b + c)^2$ is a square and $\nu_p(2013n + 2) = odd$ , then we must have $p \mid ab + bc + ca$ . Now $p \mid a + b + c$ $\Rightarrow$ $p \mid a^2 + b^2 + c^2$ and $p \mid (a+b)^2 - c^2$ $\Rightarrow$ $p \mid a^2 + b^2 + c^2 + (a+b)^2 - c^2$ $\Rightarrow$ $p \mid a^2 + b^2 + (a+b)^2$ $\Rightarrow$ $p \mid a^2 + ab + b^2$ . Now by lemma for $p \equiv 2 \pmod 3$ we have that $p \mid a$ and $p \mid b$ . Since $p \mid a + b + c$ , $p \mid c$ too $\Rightarrow$ gcd(a,b,c) = p and since we assumed gcd(a,b,c) = 1 we have a contradiction. There don't exist such a,b,c that work.

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megarnie

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megarnie

#53 h Jul 13, 2024, 4:28 PM

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No. Suppose there did exist such $a,b,c$ . Choose $a,b,c$ such that $a + b + c$ is minimal.

Claim: $\gcd(a,b,c) = 1$
Proof: If $d>1$ divided $a,b,c$ , we find that $\frac ad, \frac bd, \frac cd$ satisfies the divisibility, contradiction to the minimality of $a+b+c$ . $\square$

Now let $a^2 + b^2 + c^2 = 2013k (ab + bc + ca)$ . We find that $(2013k + 2) (ab + bc + ca) = (a+b+c)^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
Since $2013k + 2\equiv 2\pmod 3$ , there exists a prime $p\equiv 2\pmod 3$ with $\nu_p(2013k + 2)$ is odd. Choose $p$ minimal.

Case 1: $p = 2$
Then notice the LHS of $(1)$ must be a perfect square, so it has an even $\nu_2$ , meaning that $ab + bc + ca$ is even and $a + b + c$ is also even. Since $\gcd(a,b,c) = 1$ , at least one of $a,b,c$ is odd. Hence two of $a,b,c$ are odd and the other is even, but this is absurd since it means $ab + bc + ca$ is odd.

Case 2: $p > 2$
Then notice the LHS of $(1)$ must have an even $\nu_p$ , so $p$ divides $ab + bc + ca$ and $a + b + c$ . But then $p$ also divides $(a+b+c)^2 - 2(ab + bc + ca) = a^2 + b^2 + c^2$ . If we write $c \equiv -(a+b)\pmod p$ , we see that $p$ divides $a^2 + b^2 + (a+b)^2 = 2(a^2 + ab + b^2)$ . Since $p$ is odd, $p$ also divides $a^2 + ab + b^2$ . If one of $a,b$ is divisible by $p$ , then all three of $a,b,c$ are, contradicting $\gcd(a,b,c) = 1$ . Now, we find that $p$ divides $(a-b)(a^2 + ab + b^2) = a^3 - b^3$ , so this also means that $p$ divides $(a/b)^3 - 1$ . If $a/b \not \equiv 1\pmod p$ , then the order of $\frac ab$ modulo $p$ equals $3$ , so $p\equiv 1\pmod 3$ , contradiction. Hence $\frac ab \equiv 1\pmod p$ , so $a\equiv b \pmod p$ . But then $a^2 + ab + b^2 \equiv 3a^2 \pmod p$ , so $p\mid 3a^2$ . Since $p \equiv 2\pmod 3$ , $p$ cannot equal $3$ , and since $p\nmid a$ , this is absurd.

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Likeminded2017

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Likeminded2017

#54 h Oct 18, 2024, 5:00 AM

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I used way too many hints after I got stuck at the $p \equiv 3 \mod 2...$

Suppose $a^2+b^2+c^2=k(2013)(ab+bc+ca).$ Choose $a,b,c$ with minimal $a.$ We have that
$(a+b+c)^2=(2013k+2)(ab+bc+ca)$ Observe $2013k+2 \equiv 2 \mod 3$ so $\exists p \equiv 3 \mod 2$ such that $p \mid 2013k+2,$ and observe there must exist such a $p$ to an odd power or else $2013k+2$ would $\equiv 1 \mod 3$ forming a contradiction. Thus $p \equiv 2 \mod 3$ divides $(a+b+c)$ and $(ab+bc+ca).$ Next observe
$p \mid (a+b+c)(a+b)-ab-ac-bc=a^2+b^2+ab$ Next, observe
$p \equiv 3 \mod 2 \iff a^2+ab+b^2 \equiv 0 \mod p \implies a,b \equiv 0 \mod p$ as if we let $x=ab^{-1}$ we know $x^2+x+1 \equiv 0 \mod p$ and as $(x-1)(x^2+x+1)=x^3-1$ this is a cyclotomic polynomial and solutions $p$ must either be $3$ or $\equiv 1 \mod 3.$ As $p \equiv 3 \mod 2$ we know that $p \mid a,b$ and thus $p \mid a+b+c-a-b=c,$ so taking $\left( \dfrac{a}{p}, \dfrac{b}{p}, \dfrac{c}{p} \right)$ contradicts the minimality of $a$ and we are done.

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zuat.e

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zuat.e

#55 h Mar 17, 2025, 7:08 PM

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We claim those integers don't exist:
Suppose it is true, that is $a^2-2013ka(b+c)+(b^2+c^2-2013kbc)=0$ , then if $(a,b,c)$ is a solution to such equation, so will be $(\frac{b^2+c^2-2013kbc}{a},b,c)$ ; but $a^2+b^2+c^2\equiv0 \pmod{11}$ and similarly $\frac{b^2+c^2-2013kbc}{a}+b^2+c^2\equiv \frac{b^2+c^2}{a}+b^2+c^2\equiv 0$ ( $a\nmid 11$ , because in that case $-c^2\equiv a^2+b^2\equiv b^2$ , but $(\frac{-1}{11})\equiv -1$ ), so $b^2+c^2\equiv a^3$ , yet also $b^2+c^2\equiv -a^2$ , hence $a\equiv -1$ .
Similarly, $b,c\equiv -1$ , yielding $a^2+b^2+c^2\equiv -3\pmod{11}$ , a contradiction.

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