Does x^2+y^2=z^50 have integer solutions?
alvarez...@gmail.com
Robert Israel
> Does x^2+y^2=z^50 have integer solutions >0?
Hint: if z = a^2 + b^2 = (a + i b)(a - i b), try
x + i y = ...
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
victor_me...@yahoo.co.uk
> Does x^2+y^2=z^50 have integer solutions >0?
Indeed it does; what's more it even has lots
of solutions with x, y and z pairwise coprime.
alvarez...@gmail.com
where x^25 in the right
lwa...@lausd.net
> Does x^2+y^2=z^50 have integer solutions >0?
That one's easy. Just take any Pythagorean triple:
3^2+4^2 = 5^2
and multiply both sides by 5^48 to obtain:
(3^2)(5^48)+(4^2)(5^48) = 5^50
(3(5^24))^2+(4(5^24))^2 = 5^50
So x = 3(5^24), y = 4(5^24), z = 5 is a solution.
alvarez...@gmail.com
solutions >0 in integers?
Bill Dubuque
>Robert Israel <isr...@math.MyUniversitysInitials.ca>
>>alvarez...@gmail.com wrote:
>>>
>>> Does x^2+y^2 = z^50 have integer solutions >0?
>>
>> Hint: if z = a^2 + b^2 = (a + i b)(a - i b), try x + i y = ...
>
Hint 2: (cc')^n = c^n c'^n = xx + yy, where c^n = x + i y
Put z = cc', n = 50
I.e. the norm N(x) = xx' is multiplicative N(xy) = (Nx)(Ny)
so to get a norm that's an nth power, take the norm of an nth power
i.e. N(c^n) = (Nc)^n
or c^n c'^n = (cc')^n
--Bill Dubuque
Timothy Murphy
> Does x^2+y^2=z^50 have integer solutions >0?
A positive integer n is expressible in the form n = x^2 + y^2
if and only if each prime q = 3 mod 4 occurs to an even power in n.
(See eg Hardy & Wright, "An introduction to the theory of numbers".)
It follows that your equation has a solution for _every_ z.
(Well, you need to exclude some z if you want both x,y > 0.)
Bill Dubuque
>alvarez...@gmail.com wrote:
>>
>> Does x^2 + y^2 = z^50 have integer solutions >0?
>
> A positive integer n is expressible in the form n = x^2 + y^2
> if and only if each prime q = 3 mod 4 occurs to an even power in n.
> (See eg Hardy & Wright, "An introduction to the theory of numbers".)
>
> It follows that your equation has a solution for _every_ z.
That's trivially true: 0^2 + (z^25)^2 = z^50, so there's
no need to infer it from theorem in the prior paragraph.
> (Well, you need to exclude some z if you want both x,y > 0.)
Since the problem requires solutions >0, your remark doesn't help.
--Bill Dubuque
quasi
Right.
For example, trivially, z = 1 must be excluded.
Slightly less trivially, z = 2 must also be excluded.
In fact, I think any positive integer powers of 2 must be excluded.
It's not hard to show that z = p must be excluded if p is any prime
congruent to 3 mod 4.
More generally, I think all values of z such that z has an odd number
of total prime factors congruent to 3 mod 4 must be excluded.
But that is not all, no that is not all (hehe).
For example, I think z = 9 must be excluded,
More generally, if z is excluded, then any positive integer power of z
must also be excluded.
In fact, I suspect that any finite product of excluded z values yields
another excluded z value.
Perhaps some of my above claims are hallucinations?
If not, I'll make the following conjecture ...
Conjecture:
Let S = {1,2} U {p | p is a prime congruent to 3 mod 4}, and let T be
the multiplicative closure of S. Then a positive integer z is such
that the equation x^2 + y^2 = z^50 has no positive integer solutions
x,y iff z is in T.
quasi
quasi
No, forget the above post -- sorry -- it has many mistakes.
I'll rethink it.
quasi
quasi
Ok, here's the fix (I think).
This time I'll state the conjecture in the opposite way,
characterizing the values of z that work.
Conjecture:
A positive integer z is such that the equation
x^2 + y^2 = z^50
has a solution in positive integers x,y iff z can be expressed in the
form a^2 + b^2 where a,b are distinct positive integers.
quasi
quasi
Of course, there's no reason why 50 can't be replaced by any other
even positive integer. Based on that observation, here's the
conjecture ...
Conjecture:
Let n be a positive integer.
A positive integer z is such that the equation
x^2 + y^2 = z^(2n)
quasi
Nope, still not quite right -- there are lots of counterexamples.
At this point, I'll just ask it as a question ...
Question:
Let n be a positive integer.
For which values of z does the equation
x^2 + y^2 = z^(2n)
have a solution in positive integers x,y?
quasi
Robert Israel
x^2 + y^2 = t has a solution in positive integers iff
(i) the p-adic order of t is even for all primes p == 3 mod 4, and
(ii) the 2-adic order of t is odd or t is divisible by at least one
prime == 1 mod 4
(see OEIS sequence A000404).
So counterexamples to both directions of your conjecture are easy to find.
If z = 2^k where k is odd, then x^2 + y^2 = z has a solution in positive
integers, but x^2 + y^2 = z^(2n) does not.
If z = pq for primes p,q, p == 3 mod 4 and q == 1 mod 4, then
x^2 + y^2 = z has no solution in positive integers, but
x^2 + y^2 = z^(2n) does.
quasi
However in my conjecture, I specified that z should be expressible in
the form a^2 + b^2 where a,b are _distinct_ positive integers.
>integers, but x^2 + y^2 = z^(2n) does not.
Thus, one direction of my previous conjecture still holds.
>If z = pq for primes p,q, p == 3 mod 4 and q == 1 mod 4, then
>x^2 + y^2 = z has no solution in positive integers, but
>x^2 + y^2 = z^(2n) does.
A little testing reveals that all counterexamples to the other
direction have gcd(x,y) > 1.
Based on that observation, here's my revised conjecture ...
Conjecture:
Let n be a positive integer.
A positive integer z is such that the equation
x^2 + y^2 = z^(2n)
has a solution in relatively prime positive integers x,y iff z can be
expressed in the form a^2 + b^2 where a,b are distinct positive
integers.
Remark: I'm 100% sure of the above claim.
quasi
Timothy Murphy
I wasn't purporting to give a complete answer to the question.
I was just pointing out that it reduces to the question
of which n are expressible in the form x^2 + y^2.
I mentioned Hardy & Wright,
reference to which would answer the question completely.
If you want a complete answer, n is expressible as x^2 + y^2
with x,y > 0 iff each prime q = 3 mod 4 occurs to an even power,
and some prime not = 3 mod 4 occurs in n.
So the problem is soluble with any non-zero z
divisible by some prime p not = 3 mod 4.
no comment
> On Thu, 08 Jan 2009 20:07:24 -0600, Robert Israel
>
>
>
> <isr...@math.MyUniversitysInitials.ca> wrote:
>
> >> On Thu, 08 Jan 2009 19:49:27 -0500, quasi <qu...@null.set> wrote:
>
> >> >On Thu, 08 Jan 2009 19:39:49 -0500, quasi <qu...@null.set> wrote:
>
> >> >>On Thu, 08 Jan 2009 19:32:08 -0500, quasi <qu...@null.set> wrote:
>
> >> >>>On Fri, 09 Jan 2009 00:34:36 +0100, Timothy Murphy
What about z = 45 = 9 + 36, and x^2 + y^2 = z^2 (that is, n = 1)?
quasi
By some _odd_ prime not = 3 mod 4 (i.e. p = 1 mod 4).
But it looks like that clinches the question I was trying to answer.
Thus, based on your observations, I think the following iff condition
does the trick ...
Proposition:
Let n be a positive integer.
A positive integer z is such that the equation
x^2 + y^2 = z^(2n)
has a solution in positive integers x,y iff z has a prime factor
congruent to 1 mod 4.
quasi
quasi
Oops -- so much for 100% sure.
I need to require a,b to be relatively prime as well.
Ok, try this ...
Conjecture:
Let n be a positive integer.
A positive integer z is such that the equation
x^2 + y^2 = z^(2n)
has a solution in relatively prime positive integers x,y iff z can be
expressed in the form a^2 + b^2 where a,b are distinct, relatively
prime positive integers.
Remark: Now it's fixed (but not 100% sure).
quasi
quasi
>On Thu, 8 Jan 2009 19:13:08 -0800 (PST), no comment
><adler...@gmail.com> wrote:
>
>>On Jan 8, 6:46 pm, quasi <qu...@null.set> wrote:
>>> On Thu, 08 Jan 2009 20:07:24 -0600, Robert Israel
>>>
>>> Conjecture:
>>>
>>> Let n be a positive integer.
>>>
>>> A positive integer z is such that the equation
>>>
>>> x^2 + y^2 = z^(2n)
>>>
>>> has a solution in relatively prime positive integers x,y iff z can be
>>> expressed in the form a^2 + b^2 where a,b are distinct positive
>>> integers.
>>>
>>> Remark: I'm 100% sure of the above claim.
>>>
>>> quasi
>>
>>What about z = 45 = 9 + 36, and x^2 + y^2 = z^2 (that is, n = 1)?
>
>Oops -- so much for 100% sure.
>
>I need to require a,b to be relatively prime as well.
>
>Ok, try this ...
>
>Conjecture:
>
>Let n be a positive integer.
>
>A positive integer z is such that the equation
>
> x^2 + y^2 = z^(2n)
>
>has a solution in relatively prime positive integers x,y iff z can be
>expressed in the form a^2 + b^2 where a,b are distinct, relatively
>prime positive integers.
>
>Remark: Now it's fixed (but not 100% sure).
No, still broken.
It looks like I need the even-odd condition as used for Fermat
triples.
Ok, I'll try again ...
Conjecture:
Let n be a positive integer.
A positive integer z is such that the equation
x^2 + y^2 = z^(2n)
has a solution in relatively prime positive integers x,y iff z can be
expressed in the form a^2 + b^2 where a,b are relatively
prime positive integers, with one odd, the other even.
Remark: With so many prior misses, it's hard to be fully sure, but I'd
say that for the latest version, I'm 90% sure.
quasi
Bill Dubuque
>Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>Timothy Murphy <gayl...@eircom.net> wrote:
>>>alvarez...@gmail.com wrote:
>>>>
>>>> Does x^2 + y^2 = z^50 have integer solutions >0?
>>>
>>> A positive integer n is expressible in the form n = x^2 + y^2
>>> if and only if each prime q = 3 mod 4 occurs to an even power in n.
>>> (See eg Hardy & Wright, "An introduction to the theory of numbers".)
>>>
>>> It follows that your equation has a solution for _every_ z.
>>
>> That's trivially true: 0^2 + (z^25)^2 = z^50, so there's
>> no need to infer it from theorem in the prior paragraph.
>>
>>> (Well, you need to exclude some z if you want both x,y > 0.)
>>
>> Since the problem requires solutions >0, your remark doesn't help.
>
> I wasn't purporting to give a complete answer to the question.
> I was just pointing out that it reduces to the question of which
> n are expressible in the form x^2 + y^2. I mentioned Hardy & Wright,
> reference to which would answer the question completely.
Reference? I don't recall H&W treating sums of _nonzero_ squares.
I don't see how you expect the OP to have any hope of attaining
such a reduction - esp. not merely from what you wrote above.
> If you want a complete answer, n is expressible as x^2 + y^2
> with x,y > 0 iff each prime q = 3 mod 4 occurs to an even power,
> and some prime not = 3 mod 4 occurs in n.
No, n = 1,4,16,...4^k yields infinitely many counterexamples.
--Bill Dubuque
Timothy Murphy
>> I wasn't purporting to give a complete answer to the question.
>> I was just pointing out that it reduces to the question of which
>> n are expressible in the form x^2 + y^2. I mentioned Hardy & Wright,
>> reference to which would answer the question completely.
>
> Reference? I don't recall H&W treating sums of _nonzero_ squares.
> I don't see how you expect the OP to have any hope of attaining
> such a reduction - esp. not merely from what you wrote above.
Aren't you being a bit ridiculous?
It is obvious how to extend the argument in Hardy & Wright
to cover the strictly positive case.
To repeat my point: the original question
really reduces to determining when n can be expressed as x^2 + y^2.
The additional condition that x,y > 0 only makes the argument
slightly messier.
I wasn't purporting to give the OP a complete answer;
I was just pointing out the simplest way to get such an answer.
Bill Dubuque
>Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>Timothy Murphy <gayl...@eircom.net> wrote:
>>>Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>>>Timothy Murphy <gayl...@eircom.net> wrote:
>>>>>alvarez...@gmail.com wrote:
>>>>>>
>>>>>> Does x^2 + y^2 = z^50 have integer solutions >0?
>>>>>
>>>>> A positive integer n is expressible in the form n = x^2 + y^2
>>>>> if and only if each prime q = 3 mod 4 occurs to an even power in n.
>>>>> (See eg Hardy & Wright, "An introduction to the theory of numbers".)
>>>>>
>>>>
>>>> That's trivially true: 0^2 + (z^25)^2 = z^50, so there's
>>>>
>>>>> (Well, you need to exclude some z if you want both x,y > 0.)
>>>>
>>>> Since the problem requires solutions >0, your remark doesn't help.
>>>
>>> I was just pointing out that it reduces to the question of which
>>> n are expressible in the form x^2 + y^2. I mentioned Hardy & Wright,
>>> reference to which would answer the question completely.
>>
>> I don't see how you expect the OP to have any hope of attaining
>> such a reduction - esp. not merely from what you wrote above.
>>
>>> with x,y > 0 iff each prime q = 3 mod 4 occurs to an even power,
>>> and some prime not = 3 mod 4 occurs in n.
>>
>> No, n = 1,4,16,...4^k yields infinitely many counterexamples.
>
You gave hints that either didn't apply and/or contained false
statements (as have others). So _who's_ "being a bit ridiculous"?
> It is obvious how to extend the argument in Hardy & Wright
> to cover the strictly positive case.
Iirc H&W give at least 4 proofs. Which one do you think has an
"obvious" extension? If it is so "obvious" then why did your
prior answer have infinitely many counterexamples, and why
does this thread contain so many false "theorems"? Please do
elaborate on the details of this so-called "obvious" proof.
--Bill Dubuque
quasi
wrote:
I'll take credit for most of the "false theorems" in this thread. I
jumped a little too quickly (many times) -- sorry.
The positivity requirement does introduce some subtleties which are
not immediately obvious, so I agree, the known result identifying
which positive integers can be expressed as a sum of 2 squares doesn't
instantly extend to yield a characterization of all positive integers
z such that x^2 + y^2 = z^(2n) has a solution in positive integers x,y
(where n is a given positive integer, such as, for example n=25). It's
not that the desired extension is hard, but rather, it requires a lot
of careful attention to details.
However, based on Timothy Murphy's slightly flawed specification (he
forgot to exclude the prime 2), I was able to finally achieve the
desired characterization.
I gave 2 characterizations, one for the case where x,y are required to
be relatively prime, and one for the general case.
For reference, I'll restate them ...
Proposition (1)
Let n be a positive integer.
A positive integer z is such that the equation
x^2 + y^2 = z^(2n)
has a solution in relatively prime positive integers x,y iff z is odd
and can be expressed in the form a^2 + b^2 where a,b are relatively
prime positive integers.
Proposition (2):
Let n be a positive integer.
A positive integer z is such that the equation
x^2 + y^2 = z^(2n)
has a solution in positive integers x,y iff z has a prime factor
congruent to 1 mod 4.
Remark:
I can visualize proofs for both of the above results, and I see both
proofs as elementary but tedious with respect to details. The proofs I
have in mind require care, and I won't have time to write out careful
proofs for the next few days. Maybe the details are not as bad as I
think they are, or maybe there's a simpler way. Others are welcome to
try.
quasi
Timothy Murphy
>> It is obvious how to extend the argument in Hardy & Wright
>> to cover the strictly positive case.
>
> Iirc H&W give at least 4 proofs. Which one do you think has an
> "obvious" extension?
Sigh.
All of them.
If one of x,y is 0 then n is a perfect square.
If n has a factor p = 1 mod 4
then it is sufficient to give a solution for p^2.
If p = x^2 + y^2 then p^2 = (x^2 - y^2)^2 + (2xy)^2.
This seems to me trivial compared to the problem itself.
f. barnes
And with each of those infinitely many such solutions xy is a multiple
of 19 and xyz is a multiple of 101.
David Bernier
110422359737857437^2 + 276811749100242716^2 = 5^50
110422359737857437 = 3 × 79 × 1999 × 4801 × 48547199,
276811749100242716 = 2^2 × 19 × 41 × 101 × 199 × 401 × 599 × 18401
64431646909858924948087806774847687^2 + 61132413077625071791758381011357784^2 = 25^50
64431646909858924948087806774847687 = 7 × 479 × 2879 × 12799 × 4515521599 × 115488848445601
61132413077625071791758381011357784 =
2^3 × 3 × 19 × 41 × 79 × 101 × 199 × 401 × 599 × 1999 × 4801 × 18401 × 48547199
It works so far . I don't know why.
David Bernier
f. barnes
Here you go.
http://www.geocities.com/fredlb37/triples3.pdf (Theorem 1.)
The proof is still a little messy, but I'm working on it.
lui...@yahoo.com
> On 09 Jan 2009 14:17:24 -0500, Bill Dubuque <w...@nestle.csail.mit.edu>
> wrote:
>
>
>
> Proposition (1)
>
> Let n be a positive integer.
>
> A positive integer z is such that the equation
>
> x^2 + y^2 = z^(2n)
>
> has a solution in relatively prime positive integers x,y iff z is odd
> and can be expressed in the form a^2 + b^2 where a,b are relatively
> prime positive integers.
>
As x^2 + y^2 = (z^n)^2 is a pythagoric triad, then "iff z is odd and
can..."
must be changed to "iff z^n is odd and can ..."
Ludovicus
quasi
Well, certainly z must be odd.
And yes, "z^n can be expressed ...",
But I believe I proved (although I'll have to go back over my notes to
be sure) the stronger statement that z itself must have the specified
form.
Do you have a counterexample -- that is, can you find x,y,z,n in N
such that
(1) x^2 + y^2 = z^(2n)
(2) gcd(x,y) = 1
(3) z cannot be expressed in the form a^2 + b^2
where a,b in N and gcd(a,b) = 1.
?
quasi
lui...@yahoo.com
> On 09 Jan 2009 14:17:24 -0500, Bill Dubuque <w...@nestle.csail.mit.edu>
> Proposition (1)
lui...@yahoo.com
> wrote:
> Proposition (1)
>
> Let n be a positive integer.
> A positive integer z is such that the equation
>
> x^2 + y^2 = z^(2n)
>
> has a solution in relatively prime positive integers x,y iff z is odd
> and can be expressed in the form a^2 + b^2 where a,b are relatively
> prime positive integers.
As x^2 + y^2 = (z^n)^2 is pythagorean triad the phrase "x,y iff z is
odd"
must be chaged to "x,y iff z^n is odd"
Ludovicus
quasi
>On Jan 9, 4:01 pm, quasi <qu...@null.set> wrote:
>> On 09 Jan 2009 14:17:24 -0500, Bill Dubuque <w...@nestle.csail.mit.edu>
>> wrote:
>> Proposition (1)
>>
>> Let n be a positive integer.
>> A positive integer z is such that the equation
>>
>> x^2 + y^2 = z^(2n)
>>
>> has a solution in relatively prime positive integers x,y iff z is odd
>> and can be expressed in the form a^2 + b^2 where a,b are relatively
>> prime positive integers.
>
>As x^2 + y^2 = (z^n)^2 is pythagorean triad the phrase "iff z is
>odd" must be changed to "iff z^n is odd"
I don't see why.
Since z,n are positive integers, z^n is odd iff z is odd.
I want to state my conditions in terms of z, not in terms of z^n.
If you can produce a counterexample, then of course, that will force a
change, but otherwise, I prefer to stay with the statement as written.
quasi
pumpledu...@gmail.com
Quasi is, of course, correct. And an integer can be represented as the
sum
of two coprime positive integers precisely when it has no prime
divisors
congruent to 3 modulo 4, a property that z and z^n necessarily
share....
de Pumpster