Matematika wrote: Find all pairs of integers $x,y$ for which $x^3+x^2+x=y^2+y$ We shall prove that the only solutions are $(0,0)$ and $(0,-1)$ Firstly, if $x<0,$ then $x(x^2+x+1)<0$ and hence $y(y+1)<0,$ but this is impossible. $x=0$ implies $y=0$ or $y=-1$ Now, assume that $x>0$ $x(x^2+x+1)=y(y+1)$ implies $4x^3+4x^2+4x+1=(2y+1)^2$ and hence $4x^3=(2y+1)^2-(2x+1)^2=4(y-x)(x+y+1)$ So, we get $x^3=(y-x)(x+y+1).$ Now let $p$ be a prime number dividing both $y-x$ and $x+y+1.$ Since $p$ also divides $x^3,$ we get $p \mid x$ and $p \mid 2x+1.$ (Note that $2x+1=x+y+1-(y-x)$) But this means $p \mid 1$ which is non-sense. Hence $y-x$ and $x+y+1$ must be relatively prime which shows that they are both perfect cubes. Let $y-x=m^3$ and $x+y+1=n^3,$ then $mn=x$ $n^3-m^3=2x+1=2mn+1$ and since $x>0,$ we get $n>m$ $(n-m)(n^2+mn+m^2)=2mn+1$ and since $n>m,$ we get $2mn+1 \geq 2(n^2+nm+m^2)=2n^2+2mn+2m^2,$ but this means $1 \geq 2(n^2+m^2)$ and hence $m=n=0$ which contradicts with $n>m.$ (Note that $m^2+mn+n^2>0$ for all $n>m.$) So, we are done.

crazyfehmy wrote: Matematika wrote: Find all pairs of integers $x,y$ for which $x^3+x^2+x=y^2+y$ since $x>0,$ we get $n>m$ $(n-m)(n^2+mn+m^2)=2mn+1$ and since $n>m,$ we get $2mn+1 \geq 2(n^2+nm+m^2)=2n^2+2mn+2m^2,$ but this means I didn't understand where comes the 2 from ? why can't we have $n=m+1$ ?

oups_GFX wrote: crazyfehmy wrote: Matematika wrote: Find all pairs of integers $x,y$ for which $x^3+x^2+x=y^2+y$ since $x>0,$ we get $n>m$ $(n-m)(n^2+mn+m^2)=2mn+1$ and since $n>m,$ we get $2mn+1 \geq 2(n^2+nm+m^2)=2n^2+2mn+2m^2,$ but this means I didn't understand where comes the 2 from ? why can't we have $n=m+1$ ? In fact, we can approach via AM-GM. Since $n > m$, we always have $n^2+m^2 > 2mn$. Therefore, \[2mn+1=(n-m)\left(n^2+mn+m^2\right) \geq n^2+mn+m^2 > 3mn\,,\] which means $mn < 1$, and hence a contradiction.

oups_GFX wrote: crazyfehmy wrote: Matematika wrote: Find all pairs of integers $x,y$ for which $x^3+x^2+x=y^2+y$ since $x>0,$ we get $n>m$ $(n-m)(n^2+mn+m^2)=2mn+1$ and since $n>m,$ we get $2mn+1 \geq 2(n^2+nm+m^2)=2n^2+2mn+2m^2,$ but this means I didn't understand where comes the 2 from ? why can't we have $n=m+1$ ? You are right, but as Batominovski's proof, we can find a contradiction even if $n=m+1$

Matematika wrote: Find all pairs of integers $x,y$ for which $x^3+x^2+x=y^2+y$ Can you post your solution as you probably solved that problem? Thank you.

myro111 wrote: Matematika wrote: Find all pairs of integers $x,y$ for which $x^3+x^2+x=y^2+y$ Can you post your solution as you probably solved that problem? Thank you. I think he can't post a solution as he probably hasn't solved that problem because he posted it in the unsolved forum.

mathmdmb wrote: myro111 wrote: Matematika wrote: Find all pairs of integers $x,y$ for which $x^3+x^2+x=y^2+y$ Can you post your solution as you probably solved that problem? Thank you. I think he can't post a solution as he probably hasn't solved that problem because he posted it in the unsolved forum. He can because he solved all the problems in both days,at least I was told he did.

Then he should post it in the Proposed & own forum

I can post the solution. I didn't post this problem as well as others in proposed and own forum cause I was a participant at the contest and neither of this problems are either my own or proposed by me. As well we still don't have the results so I am not sure if my solution is indeed correct. And problems from contests are usually posted in unsolved problems. Anyway here is my solution we can easily prove that $ x $ isn't negative and for $x=0$ we get solutions $(0,0) ; (0,-1)$ now we shall prove that for positive $x$ we don't have any solutions. $x^3=(y-x)(y+x+1)$ from here we get $y-x=0$ implies $x=0$ so this is not the case Let $m=gcd(x,y)$ $x=mx'$ $y=my'$ $gcd(x',y')=1$ and we take $m$ and $x'$ as positive numbers so from $x^3=(y-x)(y+x+1)$ we get $m^2{x'}^3=(y'-x')(m(y'+x')+1)$ cause $gcd({x'}^3,y'-x')=1$ we have $(y'-x') | m^2$ and cause $ gcd(m^2, m(y'+x')+1)=1$ we have $m^2 | (y'-x')$ so we have $m^2=|y'-x'|$ now considering two cases $y'-x'>0$ and $x'-y'<0$ we get $m^3=x'^3+2mx'+1$ in one case and $x'^3=m^3+2mx'+1$ in the other case these equalities are simetric in $x'$ and $m$ so we will solve only one of them $m^3=x'^3+2mx'+1>x'^3$ so it must be $x'^3+2mx'+1=m^3 \geq {(x'+1)}^3$ or $m \geq \frac{3}{2}(x'+1)$ and from $x'^3+1=m(m^2-2x')\geq \frac{3}{2}(x'+1)({(\frac{3}{2}(x'+1))}^2-2x')$ $x'^2-x'+1\geq\frac{27}{8}x'^2+\frac{15}{4}x'+\frac{27}{8}$ or $0\geq\frac{19}{8}x'^2+\frac{19}{4}x'+\frac{19}{8}=\frac{19}{8}(x'+1)^2>0$ So we don't have solutions for positive $x$ and we are done.

[moderator edit: just two points: 1) please do not quote the whole post immediate before you, and 2) give any post that you like a six star rating instead of saying "nice solution".] It seems ok to me.Nice solution!

I think the title of this topic (and the other topics that contain problems from Croatian olympiad) is inappropriate, because the majority of Croatian problems originate from other (older) competitions or various books. Therefore this is not a "Croatian" problem. [moderator edit: yes, you're right. I've edited the title.]

Matematika wrote: Find all pairs of integers $x,y$ for which \[x^3+x^2+x=y^2+y.\] Solution. First we notice that $x<x^2+x+1$, $y<y+1$, as we as$$\gcd(x,x^2+x+1)=\gcd(y,y+1)=1.$$If $x\ne0$, then, from $x(x^2+x+1)=y(y+1)$ and fundamental theorem of arithmetic, it follows that$$\begin{cases} x=y \\ x^2+x+1=y+1 \end{cases}\text{or }\begin{cases} x=-y \\ x^2+x+1=-y-1 \end{cases},$$which implies that $x=0$ or $x^2=-2$, contradictions. Therefore, $(x,y)=(0,0)$ or $(0,-1)$. $\blacksquare$ Note. Thanks to @nikolapavlovic for pointing out that the above proof is invalid.

ytChen wrote: and undamental theorem of arithmetic, it follows that$$\begin{cases} x=y \\ x^2+x+1=y+1 \end{cases}\text{or }\begin{cases} x=-y \\ x^2+x+1=-y-1 \end{cases},$$ Just because $(a,b)=1=(c,d)$ and $ab=cd$ doesn't imply $a=c$ and $b=d$ or $a=-c$,$b=-d$ i mean $3\cdot 10=2\cdot 15$

nikolapavlovic wrote: ytChen wrote: and undamental theorem of arithmetic, it follows that$$\begin{cases} x=y \\ x^2+x+1=y+1 \end{cases}\text{or }\begin{cases} x=-y \\ x^2+x+1=-y-1 \end{cases},$$ Just because $(a,b)=1=(c,d)$ and $ab=cd$ doesn't imply $a=c$ and $b=d$ or $a=-c$,$b=-d$ i mean $3\cdot 10=2\cdot 15$ Oops! Yeah you are right. Sorry for my stupid mistake. Thank you for pointing it out to me!