Source: 2018 Canadian Open Math Challenge Part B Problem 1 Let $(1+\sqrt2)^5 = a+b\sqrt2$, where $a$ and $b$ are positive integers. Determine the value of $a+b.$
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Tags: Comc, 2018 COMC
TuZo
06.12.2018 17:05
Just use the Newton's binomial formula!
NikoIsLife
06.12.2018 17:26
It doesn't hurt to expand. (And its probably the only way as well.)
\begin{align*}(1+\sqrt2)^5&=1+\binom51\sqrt2+\binom522+\binom532\sqrt2+\binom544+4\sqrt2\\
&=1+5\sqrt2+20+20\sqrt2+20+4\sqrt2\\
&=41+29\sqrt2\end{align*}The answer is $41+29=\boxed{70}$.
TuZo
06.12.2018 22:09
Great, this is the Newton's binomial formula!
ythomashu
06.12.2018 22:55
We can use the recursion.
$(a+b\sqrt 2)(1+\sqrt 2)=a+2b+(a+b)\sqrt 2$. $$(1,1)\to (3,2)\to(7,5)\to(17,12)\to(41,29)$$$41+29=70$
TuZo
06.12.2018 23:04
Yes, this is a smart solution!