parmenides51 wrote: Integers $a,b,c$ are such that the values of the trinomials $bx^2+cx+a$ and $cx^2+ax+b$ at $x=1234$ coincide. Can the first trinomial at $x = 1$ take the value $2009$? Let $P(x)=bx^2+cx+a$ and $Q(x)=cx^2+ax+b$ and $R(x)=P(x)-Q(x)=(b-c)x^2+(c-a)x+(a-b)$ We need $R(1234)=R(1)=0$ and $a+b+c=2009\equiv 2\pmod 3$ If $R(x)$ has degree $<2$, it must be allzero (since at least two zeroes) and so $a=b=c$ and so $a+b+c\equiv 0\pmod 3$, wrong So $R(x)$ has degree $2$ and $b-c\ne 0$ and we have, since its two roots are $1$ and $1234$ : $\frac{a-c}{b-c}=1+1234=1235$ and so $a-c=1235(b-c)$ and so again $a+b+c\equiv 0\pmod 3$ , impossible So $\boxed{\text{No}}$