No. Because $(x+2)^{2023}-x^{2023}$ would have to be perfect square to have odd number of factors. Moreover, we can get that $(x+2)^{2023}-x^{2023}\equiv 2\pmod{4}$ if $x$ is odd. Now, let $x=2^k\cdot m$ where $m$ is odd. Then $$(x+2)^{2023}-x^{2023}=2^{2023}((2^{k-1}\cdot m+1)^{2023}-2^{2023(k-1)}).$$So by looking at $v_2$, we have that $(2^{k-1}\cdot m+1)^{2023}-2^{2023(k-1)}$ must be a multiple of $2$, this is clearly impossible if $k>1$. But even if $k=1$ then $$(2^{k-1}\cdot m+1)^{2023}-2^{2023(k-1)}=(m+1)^{2022}-1$$which is odd.

Excuse me，can you please tell me what is JOM?

gnoka wrote: Excuse me，can you please tell me what is JOM? I would also like to know

samrocksnature wrote: gnoka wrote: Excuse me，can you please tell me what is JOM? I would also like to know "The JOM is a paper set by senior students for the juniors. It is usually given at the end of BIMO 2 as a final test."

Ааа, BIMO2! Ну теперь-то всё ясно, кто же BIMO2 не знает.