MHB Finding GCD of $(1+\sqrt2)^{2017}$

  • Thread starter Thread starter lfdahl
  • Start date Start date
  • Tags Tags
    Gcd
lfdahl
Gold Member
MHB
Messages
747
Reaction score
0
Find the greatest common divisor of the natural numbers, $a$ and $b$, satisfying:
$$\left(1+\sqrt{2}\right)^{2017} = a + b \sqrt{2}.$$
 
Mathematics news on Phys.org
lfdahl said:
Find the greatest common divisor of the natural numbers, $a$ and $b$, satisfying:
$$\left(1+\sqrt{2}\right)^{2017} = a + b \sqrt{2}.$$

first let us see the relationship of gcd(a,b) and gcd(c,d) where $(c+d\sqrt{2}) = (a+b\sqrt{2})(1+\sqrt{2})$
we get by mutiplication $c= a + 2b , d= a+b)$
so $gcd(c ,d) = gcd(a+2b, a + b) = gcd(a+b,b) = gcd(a,b)$
starting with power 1 we have $a= 1, b= 1$ so gcd 1
just to cross check for power 2 we have $(1+\sqrt{2})^2 = 3+ 2\sqrt{2})$ and gcd = 1
so we get the result as 1 using above rule repeatedly
 
kaliprasad said:
first let us see the relationship of gcd(a,b) and gcd(c,d) where $(c+d\sqrt{2}) = (a+b\sqrt{2})(1+\sqrt{2})$
we get by mutiplication $c= a + 2b , d= a+b)$
so $gcd(c ,d) = gcd(a+2b, a + b) = gcd(a+b,b) = gcd(a,b)$
starting with power 1 we have $a= 1, b= 1$ so gcd 1
just to cross check for power 2 we have $(1+\sqrt{2})^2 = 3+ 2\sqrt{2})$ and gcd = 1
so we get the result as 1 using above rule repeatedly
Well done, kaliprasad!
Below are two alternative solutions:

Solution I.

Note, that $(1+\sqrt{2})^{2017}(1-\sqrt{2})^{2017}=-1$.

If we expand $(1+\sqrt{2})^{2017}$ and $(1-\sqrt{2})^{2017}$ by Newton´s binomial formula, we readily see, that $(1-\sqrt{2})^{2017} = a-b\sqrt{2}$, since irrational terms of the expansions are terms with odd power of $\sqrt{2}$.
Therefore, $(1+\sqrt{2})^{2017}(1-\sqrt{2})^{2017}=(a+b\sqrt{2})(a-b\sqrt{2})=a^2-2b^2.$ Thus, $a^2-2b^2=-1$. Now, if $d$ is a greatest common divisor of $a$ and $b$, then $d|a^2$ and $d|b^2$, and therefore
$d|(a^2-2b^2)=-1$. Thus, $d=1$.

Solution II (induction).

For $n = 1,2,.. $ define the natural numbers $a_n$ and $b_n$ by $(1+\sqrt{2})^n=a_n+b_n\sqrt{2}.$ We prove, that the greatest common divisor of $a_n$ and $b_n$ is equal to $1$ for every $n = 1,2,...$.
Clearly, $a_1 = b_1 = 1$. From

\[a_{n+1}+b_{n+1}\sqrt{2}=(1+\sqrt{2})^{n+1}=(1+\sqrt{2})^n(1+\sqrt{2}) \\\\=(a_n+b_n\sqrt{2})(1+\sqrt{2}) = a_n+2b_n+(a_n+b_n)\sqrt{2}\]

it follows, that

\[a_{n+1} = a_n+2b_n, \: \: \: b_{n+1} = a_n+b_n\]

Now, any common divisor of $a_{n+1}$ and $b_{n+1}$ also divides $2b_{n+1}-a_{n+1} = a_n$ and $a_{n+1}-b_{n+1}= b_n$ and so is a common divisor of $a_n$ and $b_n$. Therefore, gcd$(a_{n+1},b_{n+1})= 1$, whenever gcd$(a_{n},b_{n})= 1$. Since, gcd$(a_1,b_1)= 1$, it follows by induction, that gcd$(a_{n},b_{n})= 1$ for all positive integers $n$. Particularly, gcd$(a,b)=$ gcd$(a_{2017},b_{2017})= 1$.
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Replies
4
Views
3K
Replies
1
Views
2K
Replies
2
Views
1K
Replies
5
Views
2K
Replies
8
Views
2K
Replies
2
Views
1K
Back
Top