Note that given \(x^{2^k}+\dfrac{1}{x^{2^k}}=n\) then \[x^{2^{k+1}}+\dfrac{1}{x^{2^{k+1}}}=\left(x^{2^k}+\dfrac{1}{x^{2^k}}\right)^2-2=n^2-2\]

Thus we can define a sequence \(\{a_k\}\) to be \(a_0=3\) and \(a_{n+1}=a_n^2-2\) which will give us \(a_k=x^{2^k}+\dfrac{1}{x^{2^k}}\)

We want to find the last two digits of \(a_{2013}\). Note that when \(k\ge 1\), the units digit of \(a_k\) is \(7\), so brute force calculation of the period of this recursion won't be that bad:
\[a_0\equiv 3\pmod{100}\]
\[a_1\equiv 7\pmod{100}\]
\[a_2\equiv 47\pmod{100}\]
\[a_3\equiv 7\pmod{100}\]
\[\vdots\]

And thus we see that the last two digits of \(a_k\) are periodic with period \(2\). It remains to find the last two digits of \(a_{2013}\), which we see is clearly \(\boxed{7}\).
–
Daniel Liu
·
2 years, 3 months ago

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@Daniel Liu
–
Actually, \(a_0 = 3\) and \(a_1 = 7\), so \(a_{2013} \equiv 7 \pmod{100}\). Also, is it possible to find an explicit formula for the sequence?
–
Pranshu Gaba
·
2 years, 3 months ago

As for an explicit formula, just note that \(x+\dfrac{1}{x}=3\iff x^2-3x+1=0\iff x=\dfrac{3\pm\sqrt{5}}{2}\)

Thus, we have \(a_n=x^{2^n}+\dfrac{1}{x^{2^n}}=\left(\dfrac{3+\sqrt{5}}{2}\right)^{2^n}+\dfrac{1}{\left(\dfrac{3+\sqrt{5}}{2}\right)^{2^n}}\)
–
Daniel Liu
·
2 years, 3 months ago

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@Daniel Liu
–
https://oeis.org/A005248 saves a lot of calculating effort :)
–
Lolly Lau
·
2 years, 1 month ago

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Isn't this one of your problems :\
–
Lolly Lau
·
2 years, 1 month ago

## Comments

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TopNewestNote that given \(x^{2^k}+\dfrac{1}{x^{2^k}}=n\) then \[x^{2^{k+1}}+\dfrac{1}{x^{2^{k+1}}}=\left(x^{2^k}+\dfrac{1}{x^{2^k}}\right)^2-2=n^2-2\]

Thus we can define a sequence \(\{a_k\}\) to be \(a_0=3\) and \(a_{n+1}=a_n^2-2\) which will give us \(a_k=x^{2^k}+\dfrac{1}{x^{2^k}}\)

We want to find the last two digits of \(a_{2013}\). Note that when \(k\ge 1\), the units digit of \(a_k\) is \(7\), so brute force calculation of the period of this recursion won't be that bad: \[a_0\equiv 3\pmod{100}\] \[a_1\equiv 7\pmod{100}\] \[a_2\equiv 47\pmod{100}\] \[a_3\equiv 7\pmod{100}\] \[\vdots\]

And thus we see that the last two digits of \(a_k\) are periodic with period \(2\). It remains to find the last two digits of \(a_{2013}\), which we see is clearly \(\boxed{7}\). – Daniel Liu · 2 years, 3 months ago

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– Pranshu Gaba · 2 years, 3 months ago

Actually, \(a_0 = 3\) and \(a_1 = 7\), so \(a_{2013} \equiv 7 \pmod{100}\). Also, is it possible to find an explicit formula for the sequence?Log in to reply

As for an explicit formula, just note that \(x+\dfrac{1}{x}=3\iff x^2-3x+1=0\iff x=\dfrac{3\pm\sqrt{5}}{2}\)

Thus, we have \(a_n=x^{2^n}+\dfrac{1}{x^{2^n}}=\left(\dfrac{3+\sqrt{5}}{2}\right)^{2^n}+\dfrac{1}{\left(\dfrac{3+\sqrt{5}}{2}\right)^{2^n}}\) – Daniel Liu · 2 years, 3 months ago

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– Lolly Lau · 2 years, 1 month ago

https://oeis.org/A005248 saves a lot of calculating effort :)Log in to reply

Isn't this one of your problems :\ – Lolly Lau · 2 years, 1 month ago

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– Benedict Dimacutac · 2 years, 1 month ago

Indeed. Thanks for your observation.Log in to reply

hmm i did the same . Answer is 47. – Sarthak Gupta · 2 years, 3 months ago

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47 – Raman Rai · 2 years, 3 months ago

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