Letting $$P_{n}$$ be the $$n$$th Pell number:

${P_{n}}^2=\frac{(-1)^{n+1}}{4}+\sum_{k=0}^{\frac{n-(n\bmod2)}{2}}{n\choose2k}\cdot\Bigg(\frac{3^n\cdot8^k}{4\cdot9^k}\Bigg)$${P_{n}}^4=\frac{3}{32}+\sum_{k=0}^{\frac{n-(n\bmod2)}{2}}{n\choose2k}\cdot\Bigg(\frac{17^n\cdot288^k}{32\cdot289^k}-\frac{(-3)^n\cdot8^k}{8\cdot9^k}\Bigg)$

Today I've been playing around with maths looking for patterns and formulae for even powers of Pell numbers.

I made a time-lapse video scrambling and solving a Large Icosaminx: youtube.com/watch?v=_qeLKs9KLH mathstodon.xyz/media/KPvL1hELH

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I made a GIF to illustrate sequence oeis.org/A028401 and I felt like playing around with that pattern… mathstodon.xyz/media/IuKBtWbSK

Last sequences I published in the OEIS are oeis.org/A282389 and oeis.org/A282390. I made a GIF that illustrates sequence A282389 so well. ☺ mathstodon.xyz/media/8Cd3kfHzP

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that $$\sqrt{2}$$ is irrational:
Suppose not, i.e. $$\sqrt{2} = \frac{a}{b}$$, $$a,b \in \mathbb{Z}$$ and $$a,b$$ are coprime.
Then $$\left(\frac{a}{b}\right)^2 = 2$$. Then $$a^2 = 2b^2$$, so $$a^2$$ is even. Let $$a = 2c$$, so we have $$4c^2 = 2b^2 \implies b^2 = 2c^2$$. So $$b$$ is even.
Then both $$a$$ and $$b$$ are divisible by 2, contradicting the earlier proposition.
So $$\sqrt{2}$$ can't be expressed as the ratio of two integers, which means it's irrational.

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