Number Triangle

The red numbers must always have an odd factor (see proof below), so $2^{2017}$ will not be colored red .

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Proof that the red numbers must always have an odd factor:

The way the numbers are written, the left side red numbers underneath the first number $n$ of each row will be $n$ rows below it. (For example, the red number $12$ is $2$ rows under the $2$ .) Note that the last number of each row is a square number, which means the first number of each row is square number plus one, or $x^2 + 1$ , so the red number underneath it is $x^2 + 1$ rows below it, making it have a value of $r_{\text{left}} = (x + (x^2 + 1))^2 + 1 + (x^2 + 1) = x^4 + 2x^3 + 4x^2 + 2x + 3$ .

By a similar argument, the right side red numbers have a value of $r_{\text{right}} = (x + (x^2 + 1))^2 + 1 + 2(x + (x^2 + 1)) + 1 - (x^2 + 1) = x^4 + 2x^3 + 4x^2 + 4x + 3$ .

Now we can consider two cases, one where $x$ is even and one where $x$ is odd:

• If $x$ is even, then both $r_{\text{left}} = x^4 + 2x^3 + 4x^2 + 2x + 3$ and $r_{\text{right}} = x^4 + 2x^3 + 4x^2 + 4x + 3$ are odd, which means they only have odd factors.

• If $x$ is odd, then let $x = 2k + 1$ for some integer $k$ , and $r_{\text{left}} = 4(4k^4 + 12k^3 + 16k^2 + 10k + 3)$ and $r_{\text{right}} = 2(8k^4 + 24k^3 + 32k^2 + 22k + 7)$ , both of which have odd factors.

Either way, a red number must always have an odd factor.

How can you possibly prove this? I thought that 3, 12, 14 are not powers of 2 and therefore ${2}^{2017}$ would not satisfy the given condition