Prove number is natural

[valid hollow]

deew
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[novel loom]

hint: Binomial Theorem

you basically do the same process, just with summation notation

instead of writing it out long form

$(1+\sqrt{2})^{2023}=\sum_{k=0}^{2023}\binom{2023}{k}(\sqrt{2})^k$ and $(1-\sqrt{2})^{2023}=\sum_{k=0}^{2023}\binom{2023}{k}(-1)^k(\sqrt{2})^k$

so the terms of them match if k is even, and cancel if k is odd

[valid hollow]

Omegabet_

[novel loom]

but when k is even, $(\sqrt2)^k$ is an integer, and $\binom{2023}{k}$ is an integer, so $(1+\sqrt{2})^{2023}+(1-\sqrt{2})^{2023}$ is a sum of integers

[valid hollow]

Omegabet_

[novel loom]

cause again, the odd k terms will cancel each other

you're not asked what the natural number is, just that it is a natural number

[uncut summit]

@crisp estuary has given 1 rep to @novel loom