Discrete Geometry and Extremal Graph Theory - IMSA

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Other topics I have explored with my mentor include the following problems. 1. Let F n be the Fibonacci sequence defined by F 1 = F 2 = 1 and F n+2 = F n+1 + F n for n ≥ 1. Show that gcd(F m , F n ) = F gcd(m,n) . Proof. First we note the following (which are easily proven by induction) • gcd(F n+1 , F n ) = 1 • F m+n = F m+1 F n + F m F n−1 = F m F n+1 + F m−1 F n • n | m ⇒ F n | F m For instance, the first two can be done with straightforward induction while the third follows from induction and setting m = (k − 1)n, where m = kn. Now, without loss of generality, m ≥ n and let m = nq + r where 0 ≤ r < n. We have gcd(F m , F n ) = gcd(F nq+r , F n ) = gcd(F nq+1 F r + F nq F r+1 , F n ) = gcd(F r , F n ) because F n | F nq and gcd(F nq+1 , F nq ) = 1. Now, we can write n = rq 1 + r 1 and continue this procedure by writing r n = r n+1 q n+2 + r n+2 where r = r 0 , n = r −1 , and 0 ≤ r n+2 < r n+1 . Eventually, we will end up at r k = gcd(F m , F n ) and r k+1 = 0 for some k, at which point we have gcd(F m , F n ) = gcd(F rk , F rk+1 ) = gcd(F gcd(m,n) , 0) = F gcd(m,n) 2. Show that ∏ p≤x p ≤ 4x where p is a prime. Proof. The base cases x ≤ 3 are easily verified. We may assume that x > 3 is an integer, as we will prove the result by induction. Note that for odd x, the even integer x + 1 clearly cannot be prime so we have ∏ p ≤ 4 x < 4 x+1 p≤x+1 p = ∏ p≤x so the result holds for x + 1 too. Now, for odd x = 2k + 1 where k > 1, note that primes k + 2 ≤ p ≤ 2k + 1 divide ( ) 2k+1 k = (2k+1)! as they divide the numerator but k!(k+1)! not the denominator. Thus, ∏ ( ) 2k + 1 p ≤ = 1 (( ) ( )) 2k + 1 2k + 1 + < 1 k 2 k k + 1 2 (1 + 1)2k+1 = 4 k k+2≤p≤2k+1 By the induction hypothesis, we have ∏ and multiplying these yields ∏ p≤k+1 p p≤k+1 ∏ k+2≤p≤2k+1 p ≤ 4 k+1 p = ∏ p≤2k+1 p ≤ 4 2k+1 4
3. Let R m (n) be the number of subsets of an n element set, with cardinality divisible by m. In other words, let ⌊n/m⌋ ∑ ( n R m (n) = . mk) Show that R m (n) = 1 m k=0 ∑ m−1 k=0 (1 + zk ) n where z is a primitive m-th root of unity. Proof. First, we note that 0 = z m − 1 = (z − 1)(z m−1 + z m−2 + · · · + 1) and z ≠ 1 so z m−1 + z m−2 + · · · + 1 = 0. Directly expanding, we have m−1 1 ∑ m k=0 (1 + z k ) n = 1 m = 1 m m−1 ∑ k=0 n∑ j=0 n∑ j=0 ( n j) z kj ( m−1 n ∑ z j) kj k=0 Now, note that ∑ m−1 k=0 zkj equals m if m | j and 0 otherwise. Indeed, if m | j, then z j = 1 and the sum is clearly m. Otherwise, let g = gcd(m, j) and l = m g > 1 so m | lj. Then, z lj = 1 so z j is a primitive l-th root of unity. In particular, z lj − 1 = 0 and z j ≠ 1 imply z j(l−1) + z j(l−2) + · · · + 1 = 0 and we have Thus, our sum becomes as claimed. R m (n) = 1 m 4. Show that |R 3 (n) − 2n 3 | ≤ 2 3 . m−1 ∑ k=0 ∑l−1 z kj = gz kj = 0 ∑ 0≤j≤n,m|j k=0 ( n m = j) ⌊n/m⌋ ∑ k=0 ( n mk) Proof. Note that by the previous problem, we have R 3 (n) = 1 3 (2n +(1+ω) n +(1+ω 2 ) n ) where ω 3 = 1 and ω ≠ 1. Thus, it suffices to prove that |(1 + ω) n + (1 + ω 2 ) n | ≤ 2. However, this follows immediately from the triangle inequality, as |1 + ω| = | − ω 2 | = 1 and |1 + ω 2 | = | − ω| = 1, so |(1 + ω) n + (1 + ω 2 )| ≤ |(1 + ω) n | + |(1 + ω 2 ) n | = |(1 + ω)| n + |(1 + ω 2 )| n = 2. Comments. This can be generalized to a (much weaker) bound ∣ R m(n) − 2n m ∣ ≤ (2 cos π m )n for m ≥ 4 5
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Discrete Geometry and Extremal Graph Theory - IMSA

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