SOME REMARKS ON THE SOLUTION OF PROBLEM 5435, PUBLISHED IN THE ISSUE February 2017 OF THE JOURNAL
ANNA V. TOMOVA
[email protected] As shown in the figure - below, in the issue February 2017 of SSMA journal under number 5435, the following problem is published.
STAGES (PARTS) OF THE SOLUTION OF PROBLEM 5435 FIRST WAY Today, in the field of rapid development of information technology, will offer a solution to this relatively difficult problem (of this type are offered problems of the international mathematical competitions), using mathematical Internet sites and, mainly, with the help of the site http://www.wolframalpha.com/.
Basic ideas and stages of the solution: we start with so on... a method of ascent.
We find initially multiple natural values, which occupies a fraction under natural values of its two arguments with the help of the site http://www.wolframalpha.com/.
We use these values of the fraction to solve Diophantine equation with two unknowns with the help of the site http://www.wolframalpha.com/.
It turns out that this Diophantine equation with a fixed, but a specific right side always has exactly two pairs of natural numbers - its solutions. These couples a , b of solutions have very interesting properties, but upon them we will not stop. We can give and geometric explanation why this equation at fixed values b, n of the numbers has exactly two real solutions for a , not 0 or 4, but we will not deal with it. With the new ordered pair of natural solutions a , b we find new (at - large) value on the right
a 4 3a 2 1 side of the issue Diophantine equation. ab 1
We use this new value of the fraction to solve Diophantine equation with two unknowns. We get new ordered pair of natural solutions with - great value and so on... Thus we have a relatively large volume of values for numbers such that the expression of these
a 4 3a 2 1 also adopts natural value. ab 1 It turns out that for the set for the number a and b is the same. Each pair of adjacent integers of values
this set are solutions to the problem. We seek and find recurrent relationship between its members (natural numbers). This is a very important moment of the solution: finding a recurrent relationship became the appointment of mathematical site http://www.wolframalpha.com/so on... Generating function. To simplify the calculations, even with the help of computer algebra systems, we look at two sequences for recurrent arguments fraction with even and odd index in the same characteristic equation, but with different initial members. This way we find multitudes (they are endless) of natural values a , b such that expression
a 4 3a 2 1 also receives natural value. ab 1
Now is to prove the last fact. We will not use so on... a method of infinite descent but one another idea. We will try to find and find recurrent sequences of natural numbers with a
a 4 3a 2 1 in 2 cases. This way we show that ab 1 all values of the fraction after finding two sets natural numbers a , b really accept natural value. generally member - natural value of the fraction
The proof is completed. Finding of recurrent depending again became the appointment of mathematical site http://www.wolframalpha.com/ of the generating function, but after one introduction of relatively large variety of natural values of the fraction in 2 cases. Finally, we make some verifications: finding series of Taylor, some tables with the formulas and so on… Not only we find what infinite set of natural numbers, but also indicate all natural values fraction in these values arguments as members of recurrent sequence of integers. SECOND WAY
The beginning is the same, but in this second way we will prove one presentation of the fraction:
f n
4
3 f n 1 2
f n f n 1 1
f n 1 f n 1 , which shows interesting properties of this set of
positive
integers.
f n
30 7 15 4 15 30
f n
15 2 15 4 15 30
As
a
consequence
correlations
f n
4
of
n
n
Here
n 30 7 15 4 15 or 30
n 15 2 15 4 15 . 30
the
3 f n 1
latest
2
f n f n 1 1
submission
f n 1
4
of
the
3 f n 1 1
fraction,
we
get
2
f n 1 f n 2 1
. This allowing us to
choose for solutions to the problem in each of four consecutive terms of the sequence in such a way that after a suitable replacement in the numerator and denominator of the fraction it gets one and the same natural value, namely above in 2 cases. We get also the correlation:
f n
4
3 f n 1 2
f n f n 1 1
f n n N
4
3 f n 1 2
f n f n 1 1
m N .
SOLUTION OF PROBLEM 5435 FIRST WAY We find initially multiple natural values, which occupies a fraction under natural values of its two arguments with the help of the site http://www.wolframalpha.com/. Today, in the era of modern information tehnolgii we begin our attempts to solve the problem in 5435 computational verification by mathematical site http://www.wolframalpha.com/., what are the possible natural values of fraction
a 4 3a 2 1 . The results presented in the table - below. ab 1
Using a mathematical site http://www.wolframalpha.com we solve diophantine equations for several initial values of the fraction
a 4 3a 2 1 and present solutions in the following figures. ab 1
Now we change the places of two arguments and calculate new, more - great value of the fraction n
For
a 4 3a 2 1 . ab 1
a 4 3a 2 1 1077809 we obtain new solutions of Diophantine equation. ab 1
Finally, we will prove that the natural values that take two arguments of the fraction, when it is a positive integer, are:
We
divide
2
polynomial
in
ascending
degrees
of
the
variable:
P z 1 2 z 2 z z ; Q z 1 8z z . As indicated in the literature, the sequence of 2
3
2
4
coefficients of the result u1 D0 , u2 D1 ,..., un Dn 1 ,... is a recurrent sequence, the members of which
satisfy
the
equation:
Dn k
B B1 Dn k 1 ... k Dn n l k 1 . B0 B0
Here k 4, B0 1, B1 0, B2 8, B3 0, B4 1, l 3 . Therefore
Dn k
B B1 B B Dn k 1 2 Dn k 2 3 Dn k 3 4 Dn n 3 4 1 0 B0 B0 B0 B0
Dn 4 8 Dn 2 Dn n 0 un 5 8un 3 un 1 n 0 Characteristic equation has the form:
2 q 4 8q 2 1 q 4 8q 2 1 0 q1,2 4 16 1 4 15 q1,2 4 15 ; q3,4 4 15
As the four roots of the characteristic equation are different and real, then the formula for general member of the recurrent sequence has the form:
Dn A
4 15
n
B 4 15
n
C
4 15
n
D 4 15
n 0 n
. To find four
unknowns perform division in ascending degrees of variable determining the first 4 ratios.
1 A B C D
4 15 D 4 15 4 15 B 4 15 C 4 15 D 4 15 15 A 4 15 B 4 15 C 4 15 D 4 15 6 A 2 A
4 15 B 4 15 C 2
2
3
2
3
2
3
3
This is a linear algebraic system of 4 equations with 4 unknowns. Its solution is: x1 = -(sqrt(sqrt(15)+4)(2sqrt(15)-15)-2sqrt(15)-15)/60, x2 = (sqrt(sqrt(15)+4)(2sqrt(15)-15)+2sqrt(15)+15)/60, x3 = (sqrt(4-sqrt(15))(2sqrt(15)+15)-2sqrt(15)+15)/60, x4 = -(sqrt(4-sqrt(15))(2sqrt(15)+15)+2sqrt(15)-15)/60 }. or:
A
2
2
C
15 15
15 15
15 4 2 15 15 60
15 4 2 15 15
;B
2
;D
60
15 15
2
15 4 2 15 15
15 15
;
60
15 4 2 15 15 60
Then the formula for the generally member of recurrent sequence has the form: un 1 Dn
2
2
2
15 15 15 15
15 15
60
2
15 15
15 4 2 15 15 60
15 4 2 15 15 60
4 15
4 15
15 4 4 15
15 4 60
4 15
n
n
n
2
n
15 15
2 15 15
15 4 2 15 15
60
15 4 2 15 15 60
n 2 15 15 4 15 4 15 60
4 15
n
4 15
2 15 15 60
4 15
n
n
n
4 15
4 15
n
n
n 4 15
The following figure shows a relatively large number of natural values of the parameters as factors in the development of so on.. Generating function in order as in the neighborhood of 0 and in the neighborhood of infinity.
These formulas are simplified significantly in cases where the degree is odd or even.
n 2k u2 k 1 D2 k 1 2
4 15
1 4 15 2
k
4 15
2k
2 15 15 30
4 15
2k 1 4 15 4 15 15 k k k 1 4 15 4 15 4 15 15 2k
4 15
u2 k 1 D2 k
u2 k 2 D2 k 1
2k
2 15 15 30
2 15 15 4 15 30 n 2k 1
2 15 15 30
k
4 15
2k
2 15 15 30
2 15 15 4 15 30
30 7 15 4 15 30
k
2k
2k
4 15
2k
k
30 7 15 4 15 30
k
The following figures show the first 11 members of a subset of the main recurrent sequences with odd and even index.
For the first recurrent sequence the formulae for the general term is: uk 1 8uk uk 1 k 2 .
The formulae for the general term of the second sequence is the same: uk 1 8uk uk 1 k 2 . The following figures show the information for the first recurrent number obtained by mathematical site http://www.wolframalpha.com/.
The following figures show the information for the second recurrent sequence, obtained by mathematical site http://www.wolframalpha.com/.
We will explore two sequences with the same formulae to the general member but with different natural values of their first 2 members. Methodology set out - above formula will look for common member of two sequences. Characteristic equation has the form: 2 2 q 8q 1 q 8q 1 0 q1,2 4 16 1 4 15 . As the two roots of the characteristic equation are different and real, then the formula for general member of the recurrent sequences has
the form: uk A 4 15
k 1
B 4 15
k 1
k 3 . To find two unknowns perform division in
ascending degrees of variable determining the first 2 ratio.
1 A B
6 A 4 15 B 4 15
This is a linear algebraic system of 2 equations with 2 unknowns. Its solution is:
{ x = (2sqrt(15)+15)/30, y = -(2sqrt(15)-15)/30 } or A
15 2 15 15 2 15 ,B . 30 30
Then the formula for the generally member of the first recurrent sequence has the form:
uk
15 2 15 4 15 30
k 1
15 2 15 4 15 30
k 1
k 1
For the second sequence:
2 A B
15 A 4 15 B 4 15
Its
solution is: {x= (7sqrt(15)+30)/30, y = -(7sqrt(15)-30)/30 } or: 30 7 15 30 7 15 A ,B . Then the formula for the generally member of the second 30 30 recurrent sequence has the form: uk
30 7 15 4 15 30
k 1
30 7 15 4 15 30
k 1
k 1
The figures - below show results obtained using mathematical site http://www.wolframalpha.com/. For
uk
the
first
15 2 15 4 15 30
k 1
sequences
15 2 15 4 15 30
with
k 1
k 1 :
the
general
term
For
uk
the
second
30 7 15 4 15 30
k 1
sequences
30 7 15 4 15 30
with
k 1
k 1 :
the
general
term:
Finally, responding to a question 5435 is: the sets of ordered pairs of natural numbers a, b , for which the value of the fraction
a 4 3a 2 1 is natural numbers, are endless. Some of them, found by ab 1
using the mathematical site http://www.wolframalpha.com/ are shown in the above two tables. But we found a formula for general members of two recurrent sequences that contain all members of these sets (see. On - above). Finally, the natural values of the fraction
a 4 3a 2 1 are: ab 1
15 2 15 4 15 30 k 1 15 2 15 15 2 15 4 15 4 15 30 30
k 1
k 1
15 2 15 4 15 30
15 2 15 4 15 30
k 1
k
4
k 15 2 15 4 15 1 30
2
k 1 k 1 15 2 15 15 2 15 3 4 15 4 15 30 30 k 1 k 1 15 2 15 k k 15 2 15 15 2 15 15 2 15 4 15 4 15 4 15 4 15 1 30 30 30 30 1 k 1 k 1 k 1 k k 15 2 15 15 2 15 15 2 15 15 2 15 4 15 4 15 4 15 4 15 1 30 30 30 30
4
k k 15 2 15 15 2 15 4 15 4 15 30 30 k 1 k 1 k k 15 2 15 15 2 15 15 2 15 15 2 15 4 15 4 15 4 15 4 15 1 30 30 30 30
2
k k 15 2 15 15 2 15 3 4 15 4 15 30 30 k 1 k 1 k k 15 2 15 15 2 15 15 2 15 15 2 15 4 15 4 15 4 15 4 15 1 30 30 30 30 1 k 1 k 1 k 1 15 2 15 k k 15 2 15 15 2 15 15 2 15 4 15 4 15 4 15 4 15 1 30 30 30 30
30 7 15 4 15 30
30 7 15 4 15 30
30 7 15 4 15 30
k 1
k 1
k 1
30 7 15 4 15 30
30 7 15 4 15 30
k 1
k
4
k 30 7 15 4 15 1 30
2
k 1 k 1 30 7 15 30 7 15 3 4 15 4 15 30 30 k 1 k 1 k k 30 7 15 30 7 15 30 7 15 30 7 15 4 15 4 15 4 15 4 15 1 30 30 30 30 1 k 1 k 1 k 1 30 7 15 k k 30 7 15 30 7 15 30 7 15 4 15 4 15 4 15 4 15 1 30 30 30 30 and
4
k k 30 7 15 30 7 15 4 15 4 15 30 30 k 1 k 1 k k 30 7 15 30 7 15 30 7 15 30 7 15 4 15 4 15 4 15 4 15 1 30 30 30 30
2
k k 30 7 15 30 7 15 3 4 15 4 15 30 30 k 1 k 1 k k 30 7 15 30 7 15 30 7 15 30 7 15 4 15 4 15 4 15 4 15 1 30 30 30 30 1 k 1 k 1 k 1 k k 30 7 15 30 7 15 30 7 15 30 7 15 4 15 4 15 4 15 4 15 1 30 30 30 30
Note. Similarly we can prove that the same 2 recurrent sequences contain natural arguments of the fraction
b 4 3b 2 1 for which it takes natural value. ab 1
Remarks. The figures - below show results obtained using mathematical site http://www.wolframalpha.com/ for the same general terms of two sequences uk 8uk 1 uk 2 , but with different natural values of their first 2 members.
The results are the same – see the tables bellow.
Using a mathematical site http://www.wolframalpha.com/we will find two formulas for the natural
a 4 3a 2 1 of natural values of its arguments in the sets above as a general ab 1 a 4 3a 2 1 terms of two recurrent sequences. For the first set of natural values of the fraction , ab 1 values of the fraction
when
the
general
term
uk
15 2 15 4 15 30
un
30 7 15 4 15 30
figure bellow.
k 1
n
of
the
current
15 2 15 4 15 30
30 7 15 4 15 30
sequence
of
arguments
is
k 1
n
n 0 the generating function is shown on the
k 1 or
We
divide
2
polynomial
in
ascending
degrees
of
the
variable:
P z 1 58z 29 z ; Q z 1 63z 63 z z . As indicated in the literature, the sequence 2
2
3
of coefficients of the result u1 D0 , u2 D1 ,..., un Dn 1 ,... is a recurrent sequence, the members of
which
satisfy
the
equation:
Dn k
B B1 Dn k 1 ... k Dn n l k 1 . B0 B0
Here k 3, B0 1, B1 63, B2 63, B3 1, l 2 . Therefore
Dn k
B B1 B Dn k 1 2 Dn k 2 3 Dn k 3 n 3 3 1 1 B0 B0 B0
Dn 4 63Dn 2 63Dn 1 Dn n 1 un 5 63un 4 63un 3 un 1 n 0 The characteristic equation and its solutions are shown on the figure bellow.
Then the general member for the recurrent sequence is :
The value of the first 15 terms of the recurrent sequence for
a 4 3a 2 1 in the first case are shown ab 1
on the figure bellow.
The following results confirm our propositions. The series of Taylor for the generating function is:
For the second set of natural values of the fraction recurrent
sequence
uk
30 7 15 4 15 30
un
15 2 15 4 15 30
on the picture bellow.
k 1
n
a 4 3a 2 1 , when the general term of the ab 1
of
30 7 15 4 15 30
15 2 15 4 15 30
n
k 1
arguments
is
k 1 or
n 0 we find the generating function, shown
We
divide
2
polynomial
in
ascending
degrees
of
the
variable:
P z 1 34 z 5z ; Q z 1 63z 63z z . As indicated in the literature, the sequence of 2
2
3
coefficients of the result u1 D0 , u2 D1 ,..., un Dn 1 ,... is a recurrent sequence, the members of which
satisfy
the
equation:
Dn k
B B1 Dn k 1 ... k Dn n l k 1 . B0 B0
Here k 3, B0 1, B1 63, B2 63, B3 1, l 2 . As indicated in the literature, the sequence of coefficients of the result u1 D0 , u2 D1 ,..., un Dn 1 ,... is a recurrent sequence, the members of which
satisfy
the
equation:
Dn k
B B1 Dn k 1 ... k Dn n l k 1 . B0 B0
Here k 3, B0 1, B1 63, B2 63, B3 1, l 2 . The solutions of the characteristic equation are shown on the figure above. Then the general term for this recurrent sequence is :
The value of the first 15 terms of the recurrent sequence for
a 4 3a 2 1 in the second case are ab 1
shown on the figure bellow.
The following figure confirms the results, received above: Taylor series for the generating function.
SECOND WAY The beginning is the same, but in this second way we will prove one presentation of the fraction:
f n
4
3 f n 1 2
f n f n 1 1
f n 1 f n 1 which shows interesting properties of this
set of positive integers. For the first case: the following figures present the direct calculation of the first 20 natural values for the expression
f n f n 1 1 1
30 7 15 4 15 30
n
n 30 7 15 30 7 15 4 15 4 15 30 30
n 1
30 7 15 4 15 30
n 1
For the second case: the following figures present the direct calculation of the first 20 natural values for the expression
f n f n 1 1 1
15 2 15 4 15 30
n
n 15 2 15 15 2 15 4 15 4 15 30 30
n 1
15 2 15 4 15 30
n 1
It is evident that 2 sets are the same as above (in the first way). This way we prove that the
expression f n f n 1 1has the same formula for its general member of the recurrent sequence, because the generating function, the characteristic equations and their solutions are the same in two cases here and above. Now we will prove the formulae:
f n
4
3 f n 1 2
f n f n 1 1
these
f n 1
2
of
f n
30 7 15 4 15 30
f n
15 2 15 4 15 30
3 f n 1 1
f n 1 f n 2 1
sets
4
n
n
positive
n 30 7 15 4 15 or 30
n 15 2 15 4 15 . 30
, which shows interesting properties of integers.
Here
Proof.
We
f n
4
3 f n 1
f n 1 f n 1
f n f n 1 1
f n 1
4
3 f n 1 1
4
3 f n 1 1
4
2
f n 1 f n 1
f n 1 f n 1;
f n 1 f n 2 1
2
f n 2 f n 1 1
f n
3 f n 1
4
2
f n f n 1 1
f n 1
have:
f n
2
3 f n 1
f n 1 f n 1
f n 1
2
f n f n 1 1
4
3 f n 1 1 2
f n 2 f n 1 1
;
Then we have too (see above):
f n
4
3 f n 1 2
f n f n 1 1
f n
4
3 f n 1 2
f n 1 f n 1
f n 1 f n 1 n N f n 1 f n 1 m N ;
As a consequence of the latest submission of the fraction, we get the following correlations. This allowing us to choose for solutions to the problem in each of four consecutive terms of the sequence ... f n 2 , f n 1 , f n , f n 1 ... in such a way that after a suitable
replacement in the numerator and denominator of the fraction it gets one and the same natural value, namely above in 2 cases. We use the correlation:
f n
4
3 f n 1
f n n N
3 f n 1
f n 1
2
f n f n 1 1
f n
4
2
f n f n 1 1
4
3 f n 1 2
f n f n 1 1
4
3 f n 1 1
m N and
2
f n 1 f n 2 1
. The proof is completed.
Finally, we can propose the following formulas for the natural values of the fraction in 2 cases: f n f n 1 1 1 n n 30 7 15 30 7 15 30 7 15 4 15 4 15 4 15 30 30 30 or: f n f n 1 1 1
n 1
15 2 15 4 15 30
n 1
n
n 15 2 15 15 2 15 4 15 4 15 30 30
30 7 15 4 15 30
15 2 15 4 15 30
n 1
n 1
CONCLUSION Modern information technology in the use of mathematical websites and computer algebra systems help us noticeable numerical relationships, without which their use would be practically
impossible. Of course, they can not substitute for logical reasoning and conclusions. We have no reason to doubt the veracity of the results obtained, as long as they are correctly. For so on…”E mathematics” we can repeat the words of the founder of fractal geometry Benoit Mandelbrot: “While my hand inevitably gets old, e - mathematics (in the original - fractal geometry) is now the work of numerous researchers.
