SHORT SOLUTION TO THE 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.
SHORT SOLUTION TO THE PROBLEM 5435 FIRST WAY Problem 5435 allows the existence of detailed solutions that bring detailed numerical information about certain sets of numbers associated with the solution and obtained using mathematical sites or computer algebra systems, but here we will give a short solution to the problem.
This brief solution, and the content on the 5435 issue too are related to the interests of the author Valcho Millchev, advanced out in his book "Linear - recurrent sequences and Diophantine equations of the second order" UNIMAT SMB, Sofia, 2013 and - specifically, paragraph 2.1 "Two symmetrical Diophantine equations” (pp. 18-25 of that book). We will go by these results and will show how to get to the condition of a problem 5435. We will prove that for problem 5435 the symmetric Diophantine equation is x 2 8 xy y 2 11 . 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/. We find some initial positive integers a , b for which, after substitution in the numerator and denominator, the fraction
a 4 3a 2 1 takes natural value; see the ab 1
following table.
a
b
a 4 3a 2 1 ab 1
1
2
5
2
1
29
1
6
1
2
15
1
15
2
1769
118
929
1769
118
15
109621
929
7314
109621
929
118
6794705
7314
57583
6794705
…
…
…
6
1
281
47
370
281
47
6
17389
370
2913
17389
370
47
1077809
2913
22934
1077809
…
…
…
Arrange numbers - largest: 1, 2, 6,15, 47,118,... . We note the following relationship for the general term. f n 8 f n 2 f n 4 . To facilitate calculations, we will consider two linear recurrent sequences with the same formula f n 8 f n 1 f n 2 to the general member but with different starting 2 members. To avoid calculations, we use will directly result with the help of mathematical site http://www.wolframalpha.com/ for 2 formulas for the general members of two recurrent sequences, shown in the following figures.
Now we will prove that the members of these two recurrent sequences are searched values of the ordered pair, as well as some of their interesting properties. We will give a formula for natural values of the fraction in these natural values of the arguments. As is clear from the literature and in particular from p. 20 of the book "Linear - recurrent sequences and Diophantine equations of the second order" UNIMAT SMB, Sofia, 2013 and - specifically, paragraph 2.1 "Two symmetrical Diophantine equations" (pp. 18-25 of that book), author Valcho Millchev, the expression
f 2 n 8 f n f n 1 f 2 n 1 f 2 2 8 f 2 f 1 f 2 1 q 11 for 2 recurrent
sequences. Now we do the conversions:
f 2 n 8 f n f n 1 f 2 n 1 f 2 2 8 f 2 f 1 f 2 1 q 11 f 2 n 3 f n 1 8 f n f n 1 8 8 f n 1 f n 8 f 2 n 1
f n 3 f n f n 8 f n 1 f n 8 f n 1 2
2
2
2
8 f 3 n f n 1 8 f 2 n f 2 n f 2 n 1 f n f n 1 f n f n 1 1 f 4 n 3 f 2 n 1 8 f 2 n f n f n 1 1 f n f n 1 1
f n 8 f n f n 1 1 f n f n 1 1 f n f n 1 1 f n f n 1 1
We received extremely important formula in terms of this problem:
f 4 n 3 f 2 n 1 f n f n 1 1 f n f n 1 1 which proves that members of recurrent 2 sequences, the general member mentioned - above, really after replacing the numerator and denominator of the fraction make it an integer:
f 4 n 3 f 2 n 1 f n f n 1 1
f 4 n 3 f 2 n 1
f n f n 1 1 N
f n Now we will prove the formulae :
4
f n f n 1 1
3 f n 1 2
f n f n 1 1
Proof.
f n 1
4
f n f n 1 1 N 3 f n 1 1 2
f n 1 f n 2 1
We
f n
4
3 f n 1 2
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
f n
3 f n 1 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
have: 4
2
f n f n 1 1
f n 1
.
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
f n 1 f n 2 1
m N and
2
. 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
15 2 15 4 15 30
n
n 15 2 15 15 2 15 4 15 4 15 30 30
n 1
n 1
30 7 15 4 15 30
15 2 15 4 15 30
n 1
n 1
Now we show a large set of the values of the numbers of two recurrent sequences – solution of the problem:
and:
Respective natural values of the fraction are shown in the following two tables.
And
All
sets
are
endless.
A
special
place
occupies
the
pair a, b 1, 2 :
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.
