powers of 5
Franz Gnaedinger
but as in mathematics everything is connected
to everything, my simple methods often lead
to challenging questions. Here is one. Let
(aa + 1) / 2 be a power of 5. There is the
trivial solution a = 1 yielding 1 = 5 exp 0.
Then there are the true solutions a = 3
and a = 7
a = 3, (3x3 + 1) / 2 = 5
a = 7, (7x7 + 1) / 2 = 25 = 5x5
Is there any higher number yielding a power of 5?
Thanks in advance for an answer.
By the way, sci.lang has become a brothel,
and it seems that sci.math isn't off much better.
fishfry
<a883bf8c-d021-4fad...@w21g2000vby.googlegroups.com>,
Franz Gnaedinger <fr...@bluemail.ch> wrote:
It's true. The answer will cost you five bucks, big boy.
Greg Neill
...the same as in town. Or in sci.lang.
Prai Jei
continuum:
> My interest goes for very simple methods,
> but as in mathematics everything is connected
> to everything, my simple methods often lead
> to challenging questions. Here is one. Let
> (aa + 1) / 2 be a power of 5. There is the
> trivial solution a = 1 yielding 1 = 5 exp 0.
> Then there are the true solutions a = 3
> and a = 7
>
> a = 3, (3x3 + 1) / 2 = 5
>
> a = 7, (7x7 + 1) / 2 = 25 = 5x5
>
> Is there any higher number yielding a power of 5?
>
> Thanks in advance for an answer.
I can eliminate one third of the integers by considering digital roots.
The power of 5 (call it b) must be congruent to 0, 1, 2 or 4, modulo 6.
For b congruent to (0, 1, 2, 3, 4, 5) modulo 6:
5^b has digital root (1, 5, 7, 8, 4, 2) respectively
2 * 5^b - 1 (=? a^2) has digital root (1, 9, 4, 6, 7, 3) respectively.
Perfect squares cannot have digital roots of 6 or 3, so there is no solution
in these cases.
> By the way, sci.lang has become a brothel,
> and it seems that sci.math isn't off much better.
If you are interested in languages come over to alt.usage.english which is
still a reasonably friendly group.
--
ξ:) Proud to be curly
Interchange the alphabetic letter groups to reply
Franz Gnaedinger
Thank you for the answer, which I don't really
understand, however. Is there a number that
satisfies my equation, or not? Also I can exclude
many numbers, but it doesn't help me. Here is how
I proceeded:
(aa + 1) / 2 = p5 (a power of 5)
The number 'a' must be of the form 10n + 3
or 10n + 7 which leads to the following
equations
(10n + 3)(10n + 3) + 1 / 2 = p5
(10n + 7)(10n + 7) + 1 / 2 = p5
100nn + 60n + 10 / 2 = p5
100nn + 140n + 50 / 2 = p5
20nn + 30n + 5 = p5
20nn + 70n + 25 = p5
4nn + 6n + 1 = p5
4nn + 14n + 5 = p5
Now n in the first case must be 4 or 9
or 4 + 5m, and in the second case n must be
5m. And so on. The polynoms get ever larger,
but I don't reach an end.
Is alt.usage.english not flooded by porn
and ads? You are lucky, then.
Bart Goddard
@news.eternal-september.org:
> The power of 5 (call it b) must be congruent to 0, 1, 2 or 4, modulo 6.
>
This is false. A power of 5 is congruent to either 1 or 5 mod 6.
What do you mean by "digital root"?
--
Cheerfully resisting change since 1959.
Bart Goddard
b5a7-050...@w21g2000vby.googlegroups.com:
> My interest goes for very simple methods,
> but as in mathematics everything is connected
> to everything, my simple methods often lead
> to challenging questions. Here is one. Let
> (aa + 1) / 2 be a power of 5. There is the
> trivial solution a = 1 yielding 1 = 5 exp 0.
> Then there are the true solutions a = 3
> and a = 7
>
> a = 3, (3x3 + 1) / 2 = 5
>
> a = 7, (7x7 + 1) / 2 = 25 = 5x5
>
> Is there any higher number yielding a power of 5?
I write a^2 for aa and 5^k for 5 to the kth power.
If we write
a^2+1 =2*5^k, then we see that a must be odd, so
a=2n+1 for some integer n. Then
(2n+1)^2+1 = 2*5^k or
4n^2+4n+2= 2* 5^k or
2n^2+2n +1 = 5^k or
n^2 + (n+1)^2 = 5^k.
So you're looking for the sum of two consecutive
squares that equals a power of 5. My guess is that
the two solutions you have (1^2+2^2=5 and 3^2+4^2=25)
are the only ones.
achille
> Franz Gnaedinger <f...@bluemail.ch> wrote in news:a883bf8c-d021-4fad-
> b5a7-0500d75b2...@w21g2000vby.googlegroups.com:
No idea whether there are other positive integer
solutions for
x^2 + 1 = 2*5^k
but I'm pretty sure if there are other solutions,
the number is at most finite. The basic idea is factor
above equations over Gaussian integers Z[i] to get
x +/- i = u (1+i)(2+i)^k
=> | ((x+i)/(x-i))^2 - 1 | = | ((2+i)/(2-i))^(2k) + 1 | [*]
where u = +/- 1, or +/i is a unit over Z[i].
Let's consider what happens to both sides of [*]
when k is large.
For the LHS, x ~ sqrt(2)*sqrt(5)^k => LHS ~ sqrt(8)/sqrt(5)^k.
ie. LHS depends on k exponentially for large k.
For the RHS, it is known that for any algebraic number
z on the unit circle |z| = 1 which is not a root of unity,
the distance between z^k and -1 is bounded below polynomially
in k. ie. there exists number A, B > 0 such that
|z^k + 1 | >= A/k^B.
Apply this to RHS with z = ((2+i)/(2-i))^2, we see for large k,
RHS will be too big to match the LHS.
As a result, there are at most finite many k with satisfies [*].
Arturo Magidin
> Prai Jei <pvstownsend.zyx....@ntlworld.com> wrote in news:ifa26f$teu$1
> @news.eternal-september.org:
>
> > The power of 5 (call it b) must be congruent to 0, 1, 2 or 4, modulo 6.
>
> This is false. A power of 5 is congruent to either 1 or 5 mod 6.
>
> What do you mean by "digital root"?
The "digital root" is the congruence class modulo 9 (the sum of the
digits).
--
Arturo Magidin
Barry Schwarz
>> ?:) Proud to be curly
>>
>> Interchange the alphabetic letter groups to reply
>
>Thank you for the answer, which I don't really
>understand, however. Is there a number that
>satisfies my equation, or not? Also I can exclude
>many numbers, but it doesn't help me. Here is how
>I proceeded:
>
> (aa + 1) / 2 = p5 (a power of 5)
It might help others understand what you want if you used conventional
notation:
(a*a + 1) / 2 = 5^p
Excel shows that 1726334915 comes within 0.0000000000000025% of 5^26.
Not exact for number theory but close enough for engineering.
--
Remove del for email
The Pumpster
There are no further solutions (and a number of different ways to
prove it,
though I'm unsure if there's a very easy way). The most obvious way
would be
to note that any further solutions correspond to solutions to one of
x^2 - 2 y^2 = -1 or x^2 - 10 y^2 = -1
with y a power of 5, and to invoke some version of the Primitive
Divisor
Theorem (for recurrences such as those corresponding to Pell
equations,
this goes back more than 100 years).
de P
quasi
Can you give a reference?
I don't mind if the theorem is 100 years old, but I would prefer a
newer reference.
I tried searching for the theorem but only found results relating to
certain specific linear recurrences.
Based on those limited results, and assuming there is a theorem that
encompasses a wider class of linear recurrences, it seems that such a
theorem guarantees at least one new prime divisor for all terms
(values of y) beyond a certain early one (the 6'th?), so since the
prime 5 happens early, powers of 5 are soon blocked. Is that the
essence of the how the theorem you allude to gets applied?
quasi
achille
I find a reference! It is Carmichael's theorem
proved 97 years ago:
[Carmichael (1913)]
For the quadratic equation z^2 - P*z + Q = 0 where
P,Q are non-zero coprime integers. If the roots a, b
are distinct and real, then the Lucas sequence:
L_n(a,b) := (a^n - b^n)/(a-b)
contains primitive divisors for n > 12.
For x^2 - 2 y^2 = -1, y has the form
L_k(1+sqrt(2),1-sqrt(2)) for odd k.
L_1 => (x,y) = (1,1) => 1^2 - 2*5^0 = -1
L_3 => (x,y) = (7,5) => 7^2 - 2*5^1 = -1
A brute force check indicate there are no other
solutions for k <= 12. For k > 12, L_k contains
a prime factor differ from 5 first appeared in L_3.
For x^2 - 10 y^2 = -1, y has the form
L_k(3+sqrt(10),3-sqrt(10)) for odd k.
L_1 => (x,y) = (3,1) => 3^2 - 2*5*5^0 = -1
A brute force check indicate there are no other
solutions for k <= 12. For k > 12, L_k contains
a prime factor differ from 5 first appeared in
L_5 = 1405.
REF: Minoru Yabuta,
A Simple Proof of Carmichael's Theorem on Primitive Divisors
www.fq.math.ca/Scanned/39-5/yabuta.pdf
quasi
That's a cool theorem.
Thanks.
quasi
The Pumpster
I think Achille's reference is a good one. For a more general
situation,
the definitive result is in a paper of Bilu, Hanrot and Voutier :
Existence of primitive divisors of Lucas and Lehmer numbers,
J. reine angew. Math. 539 (2001), 13--23.
You can get a copy of this from Bilu's website.
de P
achille
BTW, I found another reference you might find
interesting. There is a generalization of
Carmichael's Theorem to complex Lucas pairs.
A Lucas pair is any pair of algebraic integers
(a,b) such that:
a+b and ab are non-zero coprime rational integers
a/b not a root of unity.
In 1999, Bilu-Hanrot-Voutier has proved following
(and more):
For any Lucas pair a,b; the Lucas numbers
defined by:
L_n(a,b) = (a^n-b^n)/(a-b)
contains primitive divisors for n > 30.
REF: Bilu, Hanrot and Voutier
Existence of primitive divisors of Lucas and Lehmer numbers
Franz Gnaedinger
>
> There are no further solutions (and a number of different ways to
> prove it,
> though I'm unsure if there's a very easy way). The most obvious way
> would be
> to note that any further solutions correspond to solutions to one of
>
> x^2 - 2 y^2 = -1 or x^2 - 10 y^2 = -1
>
> with y a power of 5, and to invoke some version of the Primitive
> Divisor
> Theorem (for recurrences such as those corresponding to Pell
> equations,
> this goes back more than 100 years).
>
> de P
Thank you for that clear answer. If you wonder
about my question, it has to do with a method
of calculating the circle I ascribe to Hemon,
probable designer of the Great Pyramid at Giza.
Picture a square grid measuring 10 by 10 royal
cubits:
. . . . . d . . . . .
. . e . . . . . c . .
. f . . . . . . . b .
. . . . . . . . . . .
. . . . . . . . . . .
g . . . . + . . . . a
. . . . . . . . . . .
. . . . . . . . . . .
. h . . . . . . . l .
. . i . . . . . k . .
. . . . . j . . . . .
The Sacred Triangle 3-4-5 defines a polygon
of a dozen corners and sides. Now proceed
to a finer grid of 50 by 50 smaller units.
There is a new triple 7-24-25 defining eight
more corners. Proceed to the still finer grid
250 by 250 still smaller units, and there is
a further triple 44-117-125 generating again
eight more corners. Proceed to the yet again
finer grid 625 by 625 still smaller units
and you'll find the new triple 336-527-625
that generates again eight more corners of
an ever rounder polygon. You can go on and on
this way. The radius of the circle is a power
of 5, while the first triples are
3-4-5 15-20-25 75-100-125 375-500-625 ...
7-24-25 35-120-125 175-600-625 ...
44-117-125 220-585-625 ...
336-527-625 ...
...
If you know a triple a-b-c and wish to find the next
one you may calculate these terms:
+- 4b +- 3a +- 3b +- 4a 5c
The first terms provide four results each. Use the
positive numbers that end on 1, 2, 3, 4, 6, 7, 8, 9
(neither 0 nor 5).
By connecting the 12, 20, 28, 36 ... points of the grid
you will obtain a sequence of polygons. Their side
lengths are whole number multiples of the square roots
of 2 or 5 or 2x5.
The square roots of 2 and 5 are easily approximated by
means of the following number patterns. Add a pair of
numbers and you obtain the number below, double the
first number of a line and you obtain the last number:
1 1 2
2 3 4
5 7 10
12 17 24
29 41 58
70 99 140
... ... ...
This number column has an exact equivalent in the
continued fraction for the square root of 2.
Now for the square root of 5. Multiply the first
number by a factor of 5. If possible divide all
numbers of a line by 2:
1 1 5
2 6 10
1 3 5
4 8 20
2 4 10
1 2 5
3 7 15
10 22 50
5 11 25
16 36 80
8 18 40
4 9 20
13 29 65
42 94 210
21 47 105
68 152 340
34 76 170
17 38 85
17 38 85
55 123 275
178 398 890
89 199 445
288 644 1440
144 322 720
72 161 360
... ... ....
The above numbers contain two golden sequences,
namely the so-called Fibonacci sequence and the
so-called Lucas sequence:
Ls 1 3 4 7 11 18 29 47 76 123 199 322 ...
Fs 1 1 2 3 5 8 13 21 34 55 89 144 ...
Here are the first four polygons:
http://www.seshat.ch/home/poly1.GIF
http://www.seshat.ch/home/poly2.GIF
http://www.seshat.ch/home/poly3.GIF
http://www.seshat.ch/home/poly4.GIF
http://www.seshat.ch/home/polya.GIF
http://www.seshat.ch/home/polyb.GIF
http://www.seshat.ch/home/polyc.GIF
http://www.seshat.ch/home/polyd.GIF
Franz Gnaedinger
the second four are corrected below. Here again
the simplest systematic method of calculating
the circumference of a circle.
http://www.seshat.ch/home/poly1a.GIF
http://www.seshat.ch/home/poly1b.GIF
http://www.seshat.ch/home/poly1c.GIF
http://www.seshat.ch/home/poly1d.GIF
Franz Gnaedinger
> http://www.seshat.ch/home/poly1.GIF
> http://www.seshat.ch/home/poly2.GIF
> http://www.seshat.ch/home/poly3.GIF
> http://www.seshat.ch/home/poly4.GIF
>
> http://www.seshat.ch/home/polyg1a.GIF
> http://www.seshat.ch/home/polyg1b.GIF
> http://www.seshat.ch/home/polyg1c.GIF
> http://www.seshat.ch/home/polyg1d.GIF
Gerry Myerson
<a883bf8c-d021-4fad...@w21g2000vby.googlegroups.com>,
Franz Gnaedinger <fr...@bluemail.ch> wrote:
> My interest goes for very simple methods,
> but as in mathematics everything is connected
> to everything, my simple methods often lead
> to challenging questions. Here is one. Let
> (aa + 1) / 2 be a power of 5. There is the
> trivial solution a = 1 yielding 1 = 5 exp 0.
> Then there are the true solutions a = 3
> and a = 7
>
> a = 3, (3x3 + 1) / 2 = 5
>
> a = 7, (7x7 + 1) / 2 = 25 = 5x5
>
> Is there any higher number yielding a power of 5?
Here's another way to do this kind of problem.
Any solution of a^2 + 1 = 2 x 5^k
is a solution of one of the three equations
a^2 + 1 = 2 y^3,
a^2 + 1 = 10 y^3, or
a^2 + 1 = 50 y^3,
with y a power of 5.
Now there is software around to find all solutions a, y to
equations of the form a^2 + r = s y^3 for fixed r, s.
So you use that software to solve the three equations,
then check to see which solutions give y as a power of 5,
and you're done.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
