nAalniaOMliO wrote: Let $n$ be a positive integer. On the blackboard all quadratic polynomials with positive integer coefficients, that do not exceed $n$, without real roots are written Find all $n$ for which the number of written polynomials is even If $ax^2+bx+c$ is on the board with $a\ne c$, then $cx^2+bx+a$ is also on the board and so we can erase both without change on the parity of remaining quadratics. It remains only quadraticc $ax^2+bx+a$ If $ax^2+bx+a$ is on the board and $ b$ is even, then $ax^2+(b-1)x+a$ is on the board too and so we can erase both without change on the parity of remaining quadratics. [edited] From here, this becomes wrong (thanks nAalniaOMliO for noticing it. Look at corrected post later in the thread.[End edited] It remains only quadratics $ax^2+bx+a$ where $ b$ is odd and $n+1\ge b+1>n$ (since then $ax^2+(b+1)x+a$ was not on the board) If $n=2k$ is even, we have no such $ b$ (since $ b$ odd $>n-1$ means $b=2k+1>n$) and so remaining number is zero, even. If $n=2k-1$ is odd, it remains quadratics $ax^2+(2k-1)x+a$ where $2(2k-1)\ge 2a>2k-1$ and so $2k-1\ge a\ge k$ And so exactly $ k$ quadratics on the board. And so the number of quadratics is even only when $n=2k$ or $n=4k-1$ Hence the answer : $\boxed{n\equiv 0,2,3\pmod 4}$

pco wrote: nAalniaOMliO wrote: Let $n$ be a positive integer. On the blackboard all quadratic polynomials with positive integer coefficients, that do not exceed $n$, without real roots are written Find all $n$ for which the number of written polynomials is even If $ax^2+bx+c$ is on the board with $a\ne c$, then $cx^2+bx+a$ is also on the board and so we can erase both without change on the parity of remaining quadratics. It remains only quadraticc $ax^2+bx+a$ If $ax^2+bx+a$ is on the board and $ b$ is even, then $ax^2+(b-1)x+a$ is on the board too and so we can erase both without change on the parity of remaining quadratics. It remains only quadratics $ax^2+bx+a$ where $ b$ is odd and $n+1\ge b+1>n$ (since then $ax^2+(b+1)x+a$ was not on the board) If $n=2k$ is even, we have no such $ b$ (since $ b$ odd $>n-1$ means $b=2k+1>n$) and so remaining number is zero, even. If $n=2k-1$ is odd, it remains quadratics $ax^2+(2k-1)x+a$ where $2(2k-1)\ge 2a>2k-1$ and so $2k-1\ge a\ge k$ And so exactly $ k$ quadratics on the board. And so the number of quadratics is even only when $n=2k$ or $n=4k-1$ Hence the answer : $\boxed{n\equiv 0,2,3\pmod 4}$ You forgot the "without real roots" part

nAalniaOMliO wrote: You forgot the "without real roots" part Not at all

pco wrote: It remains only quadratics $ax^2+bx+a$ where $ b$ is odd and $n+1\ge b+1>n$ (since then $ax^2+(b+1)x+a$ was not on the board) Also quadratics where $b$ is odd and $ax^2+(b+1)x+a$ has a real solution remain

nAalniaOMliO wrote: pco wrote: It remains only quadratics $ax^2+bx+a$ where $ b$ is odd and $n+1\ge b+1>n$ (since then $ax^2+(b+1)x+a$ was not on the board) Also quadratics where $b$ is odd and $ax^2+(b+1)x+a$ has a real solution remain No. Since we started with quadratics without real roots, it remains only quadratics without real roots. And when I counted them, I found : $0$ when $n=2k$ And $ k$ when $n=2k-1$ because we need $(2k-1)^2-4a^2<0$ (no real root) and so $a\ge k$ and since $2k-1\ge a$, we get only possibilities $a=k,k+1,,...,2k-1$ and so only $ k$ such polynomials

pco wrote: Since we started with quadratics without real roots, it remains only quadratics without real roots. I meant that $ax^2+bx+a$ can remain if it has no roots and $ax^2+(b+1)x+a$ has a real root. Specificly, $b=2a-1$

nAalniaOMliO wrote: pco wrote: Since we started with quadratics without real roots, it remains only quadratics without real roots. I meant that $ax^2+bx+a$ can remain if it has no roots and $ax^2+(b+1)x+a$ has a real root. Specificly, $b=2a-1$ Hmmmmmpfff ! It seems ... you are right. I'm sorry Thanks

nAalniaOMliO wrote: Let $n$ be a positive integer. On the blackboard all quadratic polynomials with positive integer coefficients, that do not exceed $n$, without real roots are written Find all $n$ for which the number of written polynomials is even Hopefully, this one will be OK If $n=1$, there is a unique suitable quadratic $x^2+x+1$ and so parity is "odd" If $n\ge 2$ : If $ax^2+bx+c$ is on the board with $a\ne c$, then $cx^2+bx+a$ is also on the board and so we can erase both without change on the parity of remaining quadratics. It remains only quadratics $ax^2+bx+a$ And so we must count parity of pairs of positive integers $(a,b)$ with constraints : $1\le a\le n$ $1\le b \le\min(n,2a-1)$ (we need discriminant negative) And so $\sum_{a=1}^n\min(2a-1,n)$ $=\sum_{1\le a<\frac{n+1}2}(2a-1)+\sum_{\frac{n+1}2\le a\le n}n$ $=\sum_{a=1}^{\left\lceil\frac{n-1}2\right\rceil}(2a-1)+\sum_{a=\left\lceil\frac{n+1}2\right\rceil}^nn$ $=\left\lceil\frac{n-1}2\right\rceil^2+\left\lfloor\frac{n+1}2\right\rfloor n$ If $n=4k$, this is $12k^2$, even If $n=4k+1$, this is $12k^2+6k+1$, odd If $n=4k+2$, this is $12k^2+12k+3$, odd If $n=4k+3$, this is $12k^2+18k+7$, odd Hence the answer : $\boxed{n\equiv 0\pmod 4}$