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 $a^2-ab+b^2=1\implies (a,b,d)=(1,1,22)\implies (x,y)=(22,22)$

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

It was proposed by Moldova.

[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?

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

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

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.

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 $x,y\in\mathbb{Z^+}$.

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?

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

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

Inequalities, Theorems...

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.

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 $

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.

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

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

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) $.

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.

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

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

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.

Probably similar to above posts but

AlgebraFC wrote: Probably similar to above posts but

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 $x,y\in\mathbb{Z^+}$.

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

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 $x,y\in\mathbb{Z^+}$.

Safal_db wrote: This is my one, every time consider, $u \geq v$ Re-Checked

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) $.

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

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 $b \le 6$.

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.

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 $(s,p)=(22,22).$ $\square$

Easy one

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

Let $\gcd(x,y) = g \implies x = gx_0, \ y = gy_0 \ \text{such that} \ \gcd(x_0 , y_0)=1$

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

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 $(x,y)=\boxed{(22,22)}$. $\blacksquare$

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.

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?

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)

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 $(x, y) = (22, 22), (1, 7), \text{ or } (7, 1)$.

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 $(x; y) \in \{(1;7),\:(7;1),\:(22; 22)\}$

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