On perfect numbers which are ratios of two Fibonacci numbers∗
... Suppose now that p > 5. Then 16 | Fmn /Fm . Assume first that 3 ∤ m. Since z(2) = 3 and 3 ∤ m, it follows that Fm is odd, therefore 16 | Fmn . Hence, 12 = z(16) | mn. However, since 9 divides F12 , we get that 9 | F12 | Fmn . Relation (2.2) together with the fact that p > 5 implies that N is coprime ...
... Suppose now that p > 5. Then 16 | Fmn /Fm . Assume first that 3 ∤ m. Since z(2) = 3 and 3 ∤ m, it follows that Fm is odd, therefore 16 | Fmn . Hence, 12 = z(16) | mn. However, since 9 divides F12 , we get that 9 | F12 | Fmn . Relation (2.2) together with the fact that p > 5 implies that N is coprime ...
1. (a) Find and describe all integers x such that the conditions x ≡ 9
... the book and your notes, but do not ask other people for help, and do not collaborate with anyone else on these exam problems. Do not use computers to solve any of these problems. If you need another copy of this exam you can download the pdf file at: http://faculty.uml.edu/dklain/numtest2home.pdf ...
... the book and your notes, but do not ask other people for help, and do not collaborate with anyone else on these exam problems. Do not use computers to solve any of these problems. If you need another copy of this exam you can download the pdf file at: http://faculty.uml.edu/dklain/numtest2home.pdf ...
The Farey Sequence and Its Niche(s)
... In the third he gives an example of F5 . In the final paragraph Farey writes, “I am not acquainted, whether this curious property of vulgar fractions has been before pointed out?; or whether it may admit of some easy or general demonstration?; which are points on which I should be glad to learn the ...
... In the third he gives an example of F5 . In the final paragraph Farey writes, “I am not acquainted, whether this curious property of vulgar fractions has been before pointed out?; or whether it may admit of some easy or general demonstration?; which are points on which I should be glad to learn the ...
How we solve Diophantine equations
... 2 has an integer solution if and only if it has a real solution and solutions modulo all prime powers. In other words, the obvious necessary conditions are also sufficient. We say that equations of degree 2 satisfy the Hasse principle. This reduces the decision problem to a finite computation, since ...
... 2 has an integer solution if and only if it has a real solution and solutions modulo all prime powers. In other words, the obvious necessary conditions are also sufficient. We say that equations of degree 2 satisfy the Hasse principle. This reduces the decision problem to a finite computation, since ...
THE ARITHMETIC LARGE SIEVE WITH AN APPLICATION TO THE
... . In the 1960’s Burgess gave the estimate np p 4 ...
... . In the 1960’s Burgess gave the estimate np p 4 ...
screen pdf: 1.8 Mo, 186 p. - IMJ-PRG
... On the Brahmagupta–Fermat–Pell equation The equation x2 dy 2 = ±1, where the unknowns x and y are positive integers while d is a fixed positive integer which is not a square, has been mistakenly called with the name of Pell by Euler. It was investigated by Indian mathematicians since Brahmagupta (6 ...
... On the Brahmagupta–Fermat–Pell equation The equation x2 dy 2 = ±1, where the unknowns x and y are positive integers while d is a fixed positive integer which is not a square, has been mistakenly called with the name of Pell by Euler. It was investigated by Indian mathematicians since Brahmagupta (6 ...
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SOLUTIONS TO HOMEWORK 2
... if p is prime, ap−1 ≡ 1 mod p, and since 17 is a prime with 17 − 1 = 16, we must have, 6816 ≡ 1 mod 17. Taking a power of 2 on both sides of the congruence, we get, 6832 ≡ 1 mod 17. After you have answered what is the mistake above, write down the correct number between 0 and 16 that is 6832 mod 17. ...
... if p is prime, ap−1 ≡ 1 mod p, and since 17 is a prime with 17 − 1 = 16, we must have, 6816 ≡ 1 mod 17. Taking a power of 2 on both sides of the congruence, we get, 6832 ≡ 1 mod 17. After you have answered what is the mistake above, write down the correct number between 0 and 16 that is 6832 mod 17. ...
Lehmer`s problem for polynomials with odd coefficients
... with c2 = (log 5)/4 and cm = log( m2 + 1/2) for m > 2. We provide in Theorem 2.4 a characterization of polynomials f ∈ Z[x] for which there exists a polynomial F ∈ Dp with f | F and M(f ) = M(F ), where p is a prime number. The proof in fact specifies an explicit construction for such a polynomial F ...
... with c2 = (log 5)/4 and cm = log( m2 + 1/2) for m > 2. We provide in Theorem 2.4 a characterization of polynomials f ∈ Z[x] for which there exists a polynomial F ∈ Dp with f | F and M(f ) = M(F ), where p is a prime number. The proof in fact specifies an explicit construction for such a polynomial F ...
On minimal colorings without monochromatic solutions to a linear
... equation x1 + · · · + xp−2 = (p − 1)xp−1 . For p = 5, we can say even more. Proposition 10. Each of the 4-colorings cΠ5 is minimal for x1 + x2 + x3 = 4x4 , and there are no other minimal colorings for x1 + x2 + x3 = 4x4 . ...
... equation x1 + · · · + xp−2 = (p − 1)xp−1 . For p = 5, we can say even more. Proposition 10. Each of the 4-colorings cΠ5 is minimal for x1 + x2 + x3 = 4x4 , and there are no other minimal colorings for x1 + x2 + x3 = 4x4 . ...
Fermat's Last Theorem
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two. The cases n = 1 and n = 2 were known to have infinitely many solutions. This theorem was first conjectured by Pierre de Fermat in 1637 in the margin of a copy of Arithmetica where he claimed he had a proof that was too large to fit in the margin. The first successful proof was released in 1994 by Andrew Wiles, and formally published in 1995, after 358 years of effort by mathematicians. The theretofore unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th century. It is among the most notable theorems in the history of mathematics and prior to its proof it was in the Guinness Book of World Records for ""most difficult mathematical problems"".