Balkan MO 2017,Q1

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sqing

41680 posts

sqing

#1 h May 4, 2017, 1:21 PM • 8 Y

Y by Problem_Penetrator, Promi, e2CLucas, Ankhongu, Hopeooooo, Slayka_Stolyara, Adventure10, Mango247

Find all ordered pairs of positive integers $(x, y)$ such that: $x^3+y^3=x^2+42xy+y^2.$

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rmtf1111

698 posts

rmtf1111

#2 h May 4, 2017, 1:33 PM • 14 Y

Y by guangzhou-2015, Snakes, opptoinfinity, vsathiam, MathInfinite, Centralorbit, ring_r, e2CLucas, AlastorMoody, yousseframzi, ZHEKSHEN, Wizard0001, Adventure10, Mango247

Let $d=gcd(x,y) \implies x=ad$ , $y=bd \implies$ $(d(a+b)-1)(a^2+b^2-ab)=43ab$ We deduce that $a,b$ are coprime with $a^2+b^2-ab$ . Thus we have that $a^2-ab+b^2=43\implies (a,b)=(x,y)=(1,7)=(7,1)$ or we have that $a^2-ab+b^2=1\implies (a,b,d)=(1,1,22)\implies (x,y)=(22,22)$

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EfthymiosN

353 posts

EfthymiosN

#3 h May 4, 2017, 1:38 PM • 4 Y

Y by Hamel, Adventure10, Mango247, ehuseyinyigit

Pretty easy problem... (I hope JBMO has a harder one...)

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sqing

41680 posts

sqing

#4 h May 4, 2017, 2:34 PM • 2 Y

Y by Adventure10, Mango247

It was proposed by Moldova.

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basemfouda2002

35 posts

basemfouda2002

#5 h May 4, 2017, 2:42 PM • 2 Y

Y by Adventure10, Mango247

[quote=rmtf1111or we have that $a^2-ab+b^2=1\implies (a,b,d)=(1,1,22)\implies (x,y)=(22,22)$ [/quote]

Can you please show how you solved this equation?

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EfthymiosN

353 posts

EfthymiosN

#6 h May 4, 2017, 2:44 PM • 2 Y

Y by Adventure10, Mango247

$(a-b)^2+ab=1$ $\Leftrightarrow$ $ab=(1-a+b)(1+a-b)$ ... Just use the fact that $(a,b)=1$

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EfthymiosN

353 posts

EfthymiosN

#7 h May 4, 2017, 2:49 PM • 2 Y

Y by Hamel, Adventure10

Or $a^2-ab+b^2 \ge ab$ $\Leftrightarrow$ $1 \ge ab$

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Ferid.---.

1008 posts

Ferid.---.

#8 h May 4, 2017, 2:50 PM • 8 Y

Y by knm2608, akmathworld, Promi, Ankhongu, opptoinfinity, ehuseyinyigit, Adventure10, Mango247

I think this problem is easy for BMO,but nice for JBMO.
My solution:
We know $x+y=\frac{x^2+42xy+y^2}{x^2-xy+y^2}=1+\frac{43xy}{x^2-xy+y^2}\in\mathbb{N}.$
Let $(x,y)=d,$ then $x=dp,y=dq,$ where $(p,q)=1.$
Then we must to show that $\frac{43pq}{p^2-pq+q^2}\in\mathbb{N}.$ We know $(p,p^2-pq+q^2)=(q,p^2-pq+q^2)=1.$
Then we have two case:
$i)$ $p^2-pq+q^2=1$ Then the solution is $p=1=q$ then $x=y=d$ Then $(x,y)=(22,22).$ Which is solution.
$ii)$ $p^2-pq+q^2=43$ Then the solution is $(p,q)=(1,7),(7,1).$ Then $(x,y)=(1,7),(7,1).$ Which is a solution.

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knm2608

468 posts

knm2608

#9 h May 4, 2017, 3:08 PM • 4 Y

Y by GGPiku, Sapi123, Adventure10, Mango247

Ferid.---. wrote:

I think this problem is easy for BMO,but nice for JBMO.
My solution:
We know $x+y=\frac{x^2+42xy+y^2}{x^2-xy+y^2}=1+\frac{43xy}{x^2-xy+y^2}\in\mathbb{N}.$
Let $(x,y)=d,$ then $x=dp,y=dq,$ where $(p,q)=1.$
Then we must to show that $\frac{43pq}{p^2-pq+q^2}\in\mathbb{N}.$ We know $(p,p^2-pq+q^2)=(q,p^2-pq+q^2)=1.$
Then we have two case:
$i)$ $p^2-pq+q^2=1$ Then the solution is $p=1=q$ then $x=y=d$ Then $(x,y)=(22,22).$ Which is solution.
$ii)$ $p^2-pq+q^2=43$ Then the solution is $(p,q)=(1,7),(7,1).$ Then $(x,y)=(1,7),(7,1).$ Which is a solution.

Very nice. Pretty much the way I solved it. Just note that the Diophantine $a^2-ab+b^2=43$ has also the solution $(a,b)=(6,7),(7,6)$ which of course gives no $x,y\in\mathbb{Z^+}$ .

This post has been edited 1 time. Last edited by knm2608, May 4, 2017, 3:14 PM

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basemfouda2002

35 posts

basemfouda2002

#10 h May 4, 2017, 3:16 PM • 5 Y

Y by Al-Handasa, Nurmuhammad06, Adventure10, Mango247, Atilla

I see that everyone looked at the gcd. But why? What prompted you to do so? Or is it just a normal thing in diophantine equations to write variables in terms of the gcd?

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knm2608

468 posts

knm2608

#11 h May 4, 2017, 3:18 PM • 2 Y

Y by Adventure10, Mango247

When you have no other alternative $\gcd$ is what you use.

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basemfouda2002

35 posts

basemfouda2002

#12 h May 4, 2017, 3:30 PM • 2 Y

Y by Adventure10, Mango247

knm2608 wrote:

When you have no other alternative $\gcd$ is what you use.

Are there any other techniques that one could try when normal factorization doesn't work? I don't want to be annoying... just trying to learn

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EfthymiosN

353 posts

EfthymiosN

#13 h May 4, 2017, 3:33 PM • 1 Y

Y by Adventure10

Inequalities, Theorems...

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rmtf1111

698 posts

rmtf1111

#14 h May 4, 2017, 4:36 PM • 2 Y

Y by samirka259, Adventure10

basemfouda2002 wrote:

knm2608 wrote:

When you have no other alternative $\gcd$ is what you use.

Are there any other techniques that one could try when normal factorization doesn't work? I don't want to be annoying... just trying to learn

Cardano's method for 3rd deg equations can help in this case, but it's messy. Also you can do a horrible case-bash.

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GoJensenOrGoHome

589 posts

GoJensenOrGoHome

#15 h May 4, 2017, 11:59 PM • 1 Y

Y by Adventure10

Yes case bash is horrible but viable, it's not hard to prove $LHS > RHS$ if $x,y \ge 23$ and lower bouand of solutions is also easy becouse if not $LHS < RHS$

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delegat

652 posts

delegat

#16 h May 5, 2017, 4:33 AM • 2 Y

Y by Adventure10, Mango247

Just to give more details about one of the already suggested solutions. We will prove that one of $x$ and $y$ must be less than $23$ . Namely, otherwise we would have $\frac{x^{3}}{23} \geq x^{2}$ , $\frac{y^{3}}{23} \geq y^{2}$ and
$\frac{22}{23}\left(x^{3}+y^{3}\right) \geq \frac{22}{23} \left(x^{2}y+xy^{2}\right) = \frac{22}{23}\left(x+y\right) xy \geq 44xy > 42xy$ and adding these up we get $LHS > RHS$ . It remains to discuss $22$ possibilities and each of them can be handled using further simple inequalities. Probably not the best approach for gold or silver medal seekers but safe way for many students. Reminds me a bit on JBMO 2005 problem 1.

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guangzhou-2015

778 posts

guangzhou-2015

#17 h May 5, 2017, 4:41 AM • 2 Y

Y by Adventure10, Mango247

The solution is $(x,y)=(1,7),(7,1),(22,22)$

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arulxz

449 posts

arulxz

#18 h May 5, 2017, 12:11 PM • 2 Y

Y by Adventure10, Mango247

$\frac{40xy}{x+y}$ is an integer

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AngleChasingXD

109 posts

AngleChasingXD

#19 h May 7, 2017, 9:38 AM • 3 Y

Y by Ankhongu, Adventure10, Mango247

Denote $x+y=s$ and $xy=p$ .
Now we have $s^3 - s^2=p (3s+40)$ . $(1)$
After some divisibility computations, we obtain
$3s+40$ divides $68800$ .Using that $3s+40$
is congruent with $1$ modulo $3$ and bigger than $40$ ,
and using in $(1)$ that $p\leq \frac {s^2}{4}$ ,
we deduce $3s+40$ is lower than $172$ .
Thus $s$ may take the following values:
$8,20,40$ or $44$ .
Checking the cases, we obtain the solutions
$(1,7)$ , $(7,1)$ , $(22,22)$ .

This post has been edited 1 time. Last edited by AngleChasingXD, May 7, 2017, 10:58 AM
Reason: Typo

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sbealing

308 posts

sbealing

#20 h May 8, 2017, 2:03 PM • 2 Y

Y by Adventure10, Mango247

Let $A=x+y$ and $B=x-y$ . WLOG $x \geq y \implies B \geq 0$ . The equation is equivalent to:
$\dfrac{A(A^2+3B^2)}{4}=11A^2-10B^2 \implies B^2=\dfrac{A^2 (44-A)}{3A+40}$
It is easy to see from this $A \leq 44$ . We now note the solutions $A=1,B=1$ and $A=44,B=0$ so we can let $2 \leq A \leq 43$

$(A,3A+40)=(A,40) \vert 40=2^3 \cdot 5$
$(44-A,3A+40)=(44-A,172) \vert 172=2^2 \cdot 43$
As $A \geq 2$ we have $44-A \leq 42 \implies 43 \not \vert 44-A$ so $(44-A,3A+40) \vert 2^2$

Hence it follows the only prime factors dividing $3A+40$ are $2,5$ as otherwise we would have a prime dividing the denominator but not the numerator.

Using the bound $46 \leq 3A+40 \leq (3 \dot 43)+40=169$ we see the only possible values for $3A+40$ are $50,64,80,100,125,128,160$ .
Checking $\pmod{3}$ we see $3A+40=64,100,160$ giving $A=8,20,40$ .

It is easy to check in the original equation the only solution is $A=8,B=6$ and the other two don't yield solutions. So our solutions are $(A,B)=(1,1),(44,0),(8,6)$

Now solving for $x,y$ we see $(x,y)=(1,0),(22,22),(7,1)$ . Clearly $(1,0)$ is not valid so:
$(x,y)=(7,1),(1,7),(22,22)$ and it is easy to see they all work.

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Safal_db

163 posts

Safal_db

#21 h May 30, 2017, 3:14 PM • 1 Y

Y by Adventure10

This is my one, every time consider, $u \geq v$

Attachments:

This post has been edited 2 times. Last edited by Safal_db, Sep 4, 2017, 8:30 AM
Reason: Edit 1

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hx.fcb

11 posts

hx.fcb

#22 h Jun 4, 2017, 9:18 PM • 2 Y

Y by Adventure10, Mango247

Can anyone tell us how many points were needed for gold, silver and bronze at BMO this year?

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knm2608

468 posts

knm2608

#23 h Jun 4, 2017, 9:23 PM • 4 Y

Y by ahmedAbd, adityaguharoy, Adventure10, Mango247

hx.fcb wrote:

Can anyone tell us how many points were needed for gold, silver and bronze at BMO this year?

The cutoffs:
For gold: 39/40
For silver: 31/40
For bronze: 16/40.

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AlgebraFC

512 posts

AlgebraFC

#24 h Jun 4, 2017, 10:51 PM • 3 Y

Y by Promi, Adventure10, Mango247

Probably similar to above posts but

Solution

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knm2608

468 posts

knm2608

#25 h Jun 5, 2017, 3:36 PM • 2 Y

Y by Adventure10, Mango247

AlgebraFC wrote:

Probably similar to above posts but

Solution

Note that the Diophantine $m^2-mn+n^2=43$ has also the solution $(m,n)=(6,7),(7,6)$ which of course gives no $x,y\in\mathbb{Z^+}$ .

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Problem_Penetrator

315 posts

Problem_Penetrator

#26 h Jun 5, 2017, 5:38 PM • 1 Y

Y by Adventure10

knm2608 wrote:

Very nice. Pretty much the way I solved it. Just note that the Diophantine $a^2-ab+b^2=43$ has also the solution $(a,b)=(6,7),(7,6)$ which of course gives no $x,y\in\mathbb{Z^+}$ .

knm2608 wrote:

AlgebraFC wrote:

Probably similar to above posts but

Solution

Note that the Diophantine $m^2-mn+n^2=43$ has also the solution $(m,n)=(6,7),(7,6)$ which of course gives no $x,y\in\mathbb{Z^+}$ .

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Safal_db

163 posts

Safal_db

#27 h Jun 25, 2017, 7:08 PM • 1 Y

Y by Adventure10

Safal_db wrote:

This is my one, every time consider, $u \geq v$

Re-Checked

This post has been edited 3 times. Last edited by Safal_db, Sep 4, 2017, 8:30 AM
Reason: Cool

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Tumon2001

449 posts

Tumon2001

#31 h Nov 16, 2017, 9:00 AM • 2 Y

Y by Adventure10, Mango247

It's simply G.C.D. chasing!

Solution: Assume that $x\neq y$ . Let $(x,y)=d$ . So, $dm=x$ and $dn=y$ , for some $m,m\in\mathbb{N}$ and $(m,n)=1$ . The equation now reduces to $dm^3+dn^3=m^2+mn+n^2$ . Observe that $(mn,m^2-mn+n^2)=(m+n,mn)=1$ . Since, $m+n|40mn$ and $m^2-mn+n^2|43mn$ , so, $m+n|40$ and $m^2-mn+n^2=43$ . Now, case checking shows that $(m,n)=(1,7), (7,1)$ .

Finally, if $x=y$ , then, $(x,y)=(22,22)$ . So, $(x,y)=(1,7), (7,1), (22,22)$ .

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GRCMIRACLES

141 posts

GRCMIRACLES

#32 h Jan 2, 2018, 4:51 AM • 2 Y

Y by Adventure10, Mango247

Can any one say me how you solved the equation a^2+b^2-ab=43?

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Taha1381

816 posts

Taha1381

#33 h May 22, 2018, 11:06 AM • 2 Y

Y by Adventure10, Mango247

GRCMIRACLES wrote:

Can any one say me how you solved the equation a^2+b^2-ab=43?

Inequality just take $a \ge b$ and show $b \le 6$ .

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awesomeness142857

28 posts

awesomeness142857

#34 h Sep 14, 2018, 6:06 AM • 2 Y

Y by Adventure10, Mango247

I HAVE A BASHY METHOD.

LET FOR SOME INTEGER $X + Y = K$

SUBSTITUTING IN OUR INITIAL EQUATION AND SIMPLIFYING WE GET A QUADRATIC IN Y WHICH IS

$(3K + 40)Y^2 - Y(40K + 3K^2) + (K^3-K^2) = 0$

DISCRIMINANT OF THIS EQUATION IS ON SIMPLIFYING GIVES

$(K^2)(3K + 40)(-K+44)$ WHICH SHALL BE POSITIVE

WHICH IMPLIES $K > 44$ IS NOT POSSIBLE

THEN JUST CHECK ALL CASES $K< 44$ AND $K = 44$

IN THIS WAY NO SOLUTION WILL BE MISSED OUT.

This post has been edited 1 time. Last edited by awesomeness142857, Jan 14, 2019, 5:07 PM
Reason: typo error

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Wizard_32

1566 posts

Wizard_32

#35 h Nov 10, 2018, 5:43 AM • 2 Y

Y by Adventure10, Mango247

We claim that the only possibility is $x=y=22.$
Let $s=x+y, p=xy.$ Then we have
$\begin{align} s^3-3sp=s^2+40p \Leftrightarrow p=\frac{s^2(s-1)}{3s+40} \in \mathbb{N} \end{align}$ Thus, $3s+40|s^2(s-1) \implies 3s+40|27(s^3-s^2)-9s^2(3s+40)+129s(3s+40)-1720(3s+40)=2^5 \times 5 \times 43$ We now apply three filters: if $3s+40=f,$ then

$f > 43,$ as $p>0$ implies $s>1.$
$3|f-1$
$(1)$ yields $s^2-4p \ge 0 \Leftrightarrow s \le 44.$ Thus, $f \le 3(44)+40=172.$

Hence, we get $3s+40 \in \{160,172\}.$ Find $p$ in each case and checking when $s^2-p$ is a perfect square we get the only possibility $(s,p)=(22,22).$ $\square$

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Mathsblogger

10 posts

Mathsblogger

#36 h Apr 13, 2019, 11:13 AM • 1 Y

Y by Adventure10

Easy one

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UnagiX

1 post

UnagiX

#37 h Nov 17, 2020, 6:18 AM

Y by

@ferid can you tell how did you reach to the conclusion (p,p^2-pq+q^2)=1

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MrOreoJuice

594 posts

MrOreoJuice

#38 h Mar 20, 2021, 2:06 PM • 1 Y

Y by Mango247

Let $\gcd(x,y) = g \implies x = gx_0, \ y = gy_0 \ \text{such that} \ \gcd(x_0 , y_0)=1$
$x^3+y^3=x^2+42xy+y^2$ $\implies g(x_0 + y_0)(x_0^2 - x_0y_0 + y_0^2) = x_0^2 + 42x_0y_0 + y_0^2$ Failed attempt -
Much better -
$x_0^2 - x_0y_0 + y_0^2 \mid x_0^2 + 42x_0y_0 + y_0^2$ $\implies x_0^2 - x_0y_0 + y_0^2 \mid 43x_0 y_0$ $\gcd(x_0,y_0) = 1 = \gcd(x_0,x_0^2 + y_0(y_0 - x_0)) = \gcd(x_0 , x_0^2 - x_0y_0 + y_0^2) = 1$ $\text{similarly} \ \gcd(y_0 , x_0^2 - x_0y_0 + y_0^2)=1$ $\implies x_0^2 - x_0y_0 + y_0^2 \mid 43$ $\implies x_0^2 - x_0y_0 + y_0^2 = \{1,43 \}$ Case 1 : $x_0^2 - x_0y_0 + y_0^2 = 1$ (quadratic in $x_0$ )
$\Delta = -3y_0^2 + 4 \ge 0 \implies 4 \ge 3y_0^2 \implies y_0 = 1$
$\implies x_0 = 1$
Hence we get $(x,y) = (22,22)$

Case 2 : $x_0^2 - x_0y_0 + y_0^2 = 43$ (again quadratic in $x_0$ )
$\Delta = -3y_0^2 + 172 \ge 0 \implies y_0 \le 7$
Doing some casework on each $y_0$ we get that $(x,y) = (1,7) , (7,1)$ (another pair because we could have assumed quadratic in $y_0$ )

Our solutions are $\boxed{(x,y) = (1,7) , (7,1) , (22,22)}$ .

This post has been edited 2 times. Last edited by MrOreoJuice, Mar 20, 2021, 2:09 PM

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jasperE3

11218 posts

jasperE3

#39 h Apr 22, 2021, 3:43 AM

Y by

The equation factors as $(x+y-1)(x^2-xy+y^2)=43xy$ . Let: $c=\gcd(x,y),x=ac,y=bc$ for some coprime positive integers $a,b$ . Then we have $(ac+bc-1)\left(a^2-ab+b^2\right)=43ab$ . We now see that $ab$ is coprime with $a^2-ab+b^2$ . There are only two cases to consider.

Case 1: $ac+bc-1=ab,a^2-ab+b^2=43$
We have from the second equation $(a-b)^2=43-ab$ and $(a+b)^2=43+3ab$ , so $43-ab$ and $43+3ab$ are squares. The only $ab$ for which this works are $ab\in\{7,42\}$ . If $ab=7$ then we get the two solutions $(x,y)\in\{\boxed{(1,7)},\boxed{(7,1)}\}$ . If $ab=42$ then $(a+b)^2=169\Rightarrow a+b=13$ . In the first equation, this is $13c=43$ , no additional solutions. $\blacksquare$

Case 2: $ac+bc-1=43ab,a^2-ab+b^2=1$
Easily, $(a-b)^2=1-ab$ , so $1-ab$ has to be square, thus $ab=1$ . Then $a=b$ , so $a=b=1$ . In the first equation, it is $c=22$ , so $(x,y)=\boxed{(22,22)}$ . $\blacksquare$

This post has been edited 1 time. Last edited by jasperE3, Apr 22, 2021, 3:43 AM

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Assassino9931

1243 posts

Assassino9931

#40 h Jan 7, 2022, 1:46 PM

Y by

Clearly $(1,1)$ does not work. With $s=x+y$ , $p=xy$ we have $s^3 - 3sp = s^2 + 40p$ , i.e. $p(3s+40) = s^3 - s^2$ . Now $p\leq \frac{s^2}{4}$ implies $s^2(3s+40) \geq 4(s^3 - s^2)$ , i.e. $s\leq 44$ ; also $98 = 2\cdot(3\cdot 3+40) \leq p(3s+40) = s^2(s-1)$ so $s\geq 5$ . A quick check of $s=5,6,\ldots,44$ (for each $s$ find whether $p$ is an integer - if yes, compute it and then solve $t^2 - st + p = 0$ ) shows that only $(7,1)$ , $(1,7)$ and $(22,22)$ are solutions.

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chenghaohu

70 posts

chenghaohu

#41 h Apr 29, 2023, 4:30 AM

Y by

For this problem, I substituted that x = ky, but I could not prove that k MUST be an integer (my argument is pretty dumb). Can someone show me the proper way to do that?

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BorisAngelov1

15 posts

BorisAngelov1

#42 h Nov 6, 2023, 6:47 PM • 1 Y

Y by topologicalsort

solution

This post has been edited 1 time. Last edited by BorisAngelov1, Nov 6, 2023, 6:51 PM
Reason: minor error

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zaidova

86 posts

zaidova

#43 h Jan 22, 2024, 8:03 PM • 2 Y

Y by monoditetra, Leparolelontane

x=da y=db
d³a³ + d³b³ = a²d² + 42d²ab + d²b²
d³a³ + d³b³ - d²a² - d²b² = 42d²ab
From there we can get ;
da³ + db³ - a² - b² = 42ab
For d=1;
a²(da-1) + b²(db-1) =42ab
i) a=b
a²(ad-1) + a²(da-1) = 42a²
2*(da-1)=42
da-1=21 da=22=x db=22=y (22,22)

ii) (da+db-1)(a²-ab+b²)=43ab
d(a+b) - 1=ab
a² + b² - ab=43
d(a+b)= a² + b² - 42
a=7 b=1
a=1 b=7 (for d=1) (x,y)= (7,1), (1,7)

So, there are 3 solutions;
(7,1) (1,7) (22,22)

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fearsum_fyz

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fearsum_fyz

#44 h Jun 27, 2024, 3:13 PM

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We begin by taking out the GCD. Let $\gcd{(x, y)} = d$ and $p = \frac{x}{d}, q = \frac{y}{d}$ so that $\gcd{(p, q)} = 1$ .
We get $d^3 (p^3 + q^3) = d^2 (p^2 + 42pq + q^2)$ so $d (p^3 + q^3) = p^2 + 42pq + q^2 \implies d (p + q) (p^2 + q^2 - pq) = p^2 + 42pq + q^2$
$p + q \vert p^2 + 42pq + q^2 \implies p + q \vert p^2 + 42pq + q^2 - (p^2 + 2pq + q^2) = 40pq \implies p + q \vert 40$ .
$p^2 + q^2 - pq \vert p^2 + 42pq + q^2 \implies p^2 + q^2 - pq \vert 43pq$ , but $p, q$ are coprime to $p^2 + q^2 - pq$ , implying that $p^2 + q^2 - pq = 43$ or $p^2 + q^2 - pq = 1$ .
This gives the answers $(x, y) = (22, 22), (1, 7), \text{ or } (7, 1)$ .

This post has been edited 1 time. Last edited by fearsum_fyz, Jun 27, 2024, 3:23 PM
Reason: correction

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aaroasdf13

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aaroasdf13

#45 h Jun 27, 2024, 4:34 PM

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A substitution can exploit the symmetry in $x; y$ . It leads to 41 cases to check manually, so it is less efficient than other methods shown.

Add $3xy(x+y)$ to complete the cube. On the RHS, complete the square:
$\begin{align*} (x+y)^3 &= (x+y)^2 + 3xy(x+y) + 40xy &&\left| \begin{pmatrix}u\\v\end{pmatrix} := \begin{pmatrix} 1&-1\\ 1&1 \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\in\mathbb{Z}^2,\quad v \geq 2\right.\\[.5em] % v^3 &= v^2 + \frac{3v(v^2-u^2)}{4} + 10(v^2-u^2)&&\left|v^2-u^2 = 4xy\right. \end{align*}$ Use the substitution $u, v\in\mathbb{Z}$ with $v \geq 2$ to decouple the diophantine equation. Multiply by 4:
$\begin{align*} -u^2(40+3v) &= v^3 - 44v^2 = v^2 (v-44) \end{align*}$ Since $v \geq 2$ , the LHS is non-positive, so the RHS must be as well. That is only possible if $2\leq v \leq 44$ . Checking them manually, only $v\in\{8; 44\}$ lead to integer solutions for $u$ , namely $(u; v) \in \{(\pm6;8), (0;44)\}$ .

Substituting back, all three yield integer solutions $(x; y) \in \{(1;7),\:(7;1),\:(22; 22)\}$

This post has been edited 4 times. Last edited by aaroasdf13, Jun 27, 2024, 5:12 PM

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Perelman1000000

102 posts

Perelman1000000

#46 h Jan 12, 2025, 6:49 PM

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$\boxed{ANSWER:(x,y)=(1,7),(7,1),(22,22)}$

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Burak0609

13 posts

Burak0609

#47 h Apr 7, 2025, 6:08 PM

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So $x^3+y^3=(x+y)(x^2-xy+y^2)$ . $x+y=\frac{x^2+42xy+y^2}{x^2-xy+y^2}=1+\frac{43xy}{(x^2-xy+y^2}$ . We suppose that $(x,y)=d$ , then $x=ad,y=bd$ and $(a,b)=1$ . We can see $\frac{43ab}{a^2-ab+b^2}\in\mathbb{N}$ . We can see $(a,a^2-ab+b^2)=(b,a^2-ab+b^2)=1.$
If $a^2-ab+b^2=1$ . The solution in this case is $a=b=1$ then $x=y=d$ Then $(x,y)=(22,22).$
If $a^2-ab+b^2=43$ . The solution in this case is $(a,b)=(1,7),(7,1).$ Then $(x,y)=(1,7),(7,1).$

This post has been edited 1 time. Last edited by Burak0609, Apr 7, 2025, 6:10 PM
Reason: typo

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