OK, here's the solution for Diophantus's problem of determining the solution for:
x2 + y2 = z2
(1) We know that we can assume that x,y,z are coprime. [See my previous blog for details]
(2) The second important insight is that z has to be odd.
(a) Assume the opposite that z is even.
(b) Then, there exists another value Z such that z = 2 * Z
(c) Also, z2 is then divisible by 4 since:
z2 = (2 * Z)2 = 4 * (Z2)
(d) We know that x,y must both be odd because of (1).
(e) Since they are odd, there must also exist values X,Y such that:
x = 2 * X + 1
y = 2 * Y + 1
(f) But x2 + y2 cannot be divisible by 4 since:
x2 + y2 = (2 * X + 1)2 + (2 * Y + 1)2 =
= 4X2 + 4X + 1 + 4Y2 + 4Y + 1 =
= 4[ X2 + X + Y2 + Y ] + 2
(g) So, we have a contradiction and we reject our assumption.
(3) Since z is odd, either x or y must be even since an odd number is always the sum of an odd and an even number.
(4) Let's assume x is even. The same argument will also work if y is even.
(5) Now, we know that:
x2 = z2 - y2 = (z - y)(z + y)
(6) And, z - y and z + y must be even since z,y are odd.
(7) So, we know that there must exist u,v,w such that:
x = (2u)
z + y = (2v)
z - y = (2w)
(8) Which means that:
(2u)2 = (2v)(2w) [From (5) and (7)]
(9) Dividing both sides by 4 gives us:
u2 = v * w
(10) We need 1 more insight before the solution. Here it is: v,w are coprime
(a) Assume that v,w are not coprime.
(b) Then, there exists d such that d > 1 and d divides both v,w
(c) Then d divides both v + w and v - w
(d) But:
z + y + z - y =
2v + 2w
So 2z = 2v + 2w which means that z = v + w
So d divides z
(e) And:
z + y - (z - y) = 2v - 2w
So 2y = 2v - 2w which means that y = v - w
So d divides y
(f) Which is a contradiction since z,y are coprime [by (1)].
(g) So, we reject our assumption.
(11) By the properties of coprimes, we know from (9),(10) that v,w are themselves squares (see here for proof). [For those who need a review of coprimes, here is a link.]
(12) So, there exists p,q such that:
v = p2
w = q2
(13) And, we have our solution since:
z = v + w = p2 + q2
y = v - w = p2 - q2
x = 2u = 2pq [Since u2 = vw means u = pq]
We also know that:
(a) p,q are relatively prime. [Otherwise, z,x,y would not be relatively prime]
(b) p,q are opposite parity (that is, one is odd and one is even) [Since z is odd]
(14) For sure enough:
(p2 + q2)2 = (2pq)2 + (p2 - q2)2
(15) Now, to generate our answer, we can pick any p,q we want so long as they are integers.
For example, if p = 2 and q = 1, we get:
z = (2)2 + (1)2 = 5.
y = (2)2 - (1)2 = 3.
x = 2pq = 2*2*1 = 4
(16) We can do even better than this because we know that for each x,y,z, if they have common factors, the relation still holds.
z = d[p2 + q2]
y = d[p2 - q2]
x = d[2pq]
For example, if p = 2 and q = 1 and d = 2, we get:
z = (2)(5) = 10
y = (2)(3) = 6
x = (2)(4) = 8
And sure enough, 62 + 82 = 36 + 64 = 100.
QED
What's also nice about this result is that it is not too difficult to apply Fermat's method of infinite descent and prove Fermat's Last Theorem for n=4.
Thanks for your post.
ReplyDeleteYes, that also works since:
Let p be any positive integer.
Let x = 2p+1 (so, x is odd)
So, x^2 = (2p+1)^2 = 4p^2+4p+1
So, y = (4p^2 + 4p + 1 - 1)/2 = 2p^2 + 2p
So, z = (4p^2 + 4p + 1 + 1)/2 = 2p^2 + 2p + 1
Now, z^2 = (2p^2 + 2p + 1)^2 = 4p^4 + 8p^3 + 8p^2 + 4p + 1
And y^2 + x^2 = (2p^2 + 2p)^2 + 4p^2 + 4p + 1 = 4p^4 + 8p^3 + 8p^2 + 4p + 1.
Cheers,
-Larry
Hi Jenny,
ReplyDeleteAre you sure that you have stated the problem correctly.
There are no integer solutions to the problem x^2 + y^2 = 7*z^2.
The reason for this is that any prime of the form 4n+3 that divides the sum of two pairs must itself have an even power in the prime factorization.
Unfortunately, it is impossible for 7=(4*1+3) to have an even power in the prime factorization 7*z^2.
Here is a web page that provides the proof.
Here is another proof in pdf format.
-Larry
Hi, Larry.
ReplyDeleteI discovered below equations to calculate Pythagorean triples when I was a secondary school student.
For x = odd numbers. 1, 3, 5, 7,...
y = (x^2-1)/2, z = (x^2+1)/2
Eg.
1, 0, 1
3, 4, 5
5, 12, 13
7, 24, 25
For x = even numbers which could be divided by 4. 4, 8, 12, 16,...
y = (x^2/4)-1, z = (x^2/4)+1
Eg.
4, 3, 5
8, 15, 17
12, 35, 37
16, 63, 65
It is easy to proof that those above equations are correct for all intergers.
But until now I still can't find the phythegorean triple equations for all even numbers including, 2, 6, 10, 14....
Is it impossible?
Please advise.
thanks and best regards
cf
Hi Chin Foo,
ReplyDeleteI'm not clear on your question.
A Pythagorean Triple that consists of all even numbers is 6,8,10 since
36 + 64 = 100
There are in fact an infinite number of them which can be derived from the formula in my blog.
Please let me know if I am misunderstanding your question.
Cheers,
-Larry
Good day, Larry.
ReplyDeleteSorry for not stated my question clearly.
I never consider (6,8,10), (10,24,26) are original Pythagorean Triple since it is 2*(3,4,5) and 2* (5,12,13).
My question, is it possible to have a equation to find out Pythagorean Triple for x = 6 (but y not 8 and z not 10), x = 10 (but y not 24 and z not 26)?
Please advise.
thanks and best regards
cf
Hi Chin Foo,
ReplyDeleteIf I understand your question correctly, the answer is no.
While it is easy to find an all even solution by doubling any solution you would like, to get a primitive solution, all three integers need to be relatively prime. This means that at most, only 1 of the integers can be even.
Please let me know if I misunderstood your question.
I wasn't sure if this is what you are asking or if you are asking if x=6, how many different possible solutions are there.
Cheers,
-Larry
-Larry
Thanks, Larry.
ReplyDeleteYou have answered my question.
best regards
cf
This comment has been removed by the author.
ReplyDeleteHi Larry,
ReplyDeleteFor finding the pythagorean triplet involving a given number i use this way. take the number square it and then factorize it in such a way that both factors are even or both odd.
Ex: for 8
8^2 = 64 = 1*64 (1 odd ,1 even)
2*32 (2 even)
4*16 (2 even)
2*32 implies the other two numbers are (32+2)/2 and (32-2)/2
4*16 implies (4+16)/2 and (16-4)/2
so we can find all the triplets involving the given number.
Hi Larry,
ReplyDeleteFor finding the pythagorean triplet involving a given number i use this way. take the number square it and then factorize it in such a way that both factors are even or both odd.
Ex: for 8
8^2 = 64 = 1*64 (1 odd ,1 even)
2*32 (2 even)
4*16 (2 even)
2*32 implies the other two numbers are (32+2)/2 and (32-2)/2
4*16 implies (4+16)/2 and (16-4)/2
so we can find all the triplets involving the given number.
There is an algorith to ind ALL Pythagorean triples. I found it during my play around to find Fermat's original proof of his last theorem. For some reason, the pro's are not very interested, the silence is overwhelming. For those interested, I will be happy to publish it properly here, as long as its origin is never doubted. It will also show why Fermat's last theorem (the proof) is closely coupled to the same algoritm.
ReplyDeletegreetings!
ReplyDeletecan you give me an algorithm or formula that generate any pythagorean triple leg a and b wherein both legs could still be expressed as a sum of 2 squares ?
regards
HI..., SHORT AND SIMPLE...HERE'S WHAT I CAME UP WITH:
ReplyDeletex2+y2=z2
x=3
x2=9
y=4
y2=16
WHICH GIVES YOU:
z=5
z2=25
can any one solve this problem to me.
ReplyDeleteSketch the surface model of a solid that simultaneously satisfies the following relation:
X2+y2≤Z2/4
Z ≥ 2, Z ≤ 9
Good evening, my name is Renato.I'm degree in civil engineering. I found an interesting property about the primes. I wonder if the property is existing in the literature or not.
ReplyDeletep is a prime number if there is only one solution to the equation, x and y being integers. (The "trivial" solution is such that x + y = p and y = x + 1)
The equation is:
y² - x² = p
Thank you
(a+b)^2-2ab=a^2+b^2
ReplyDelete(3+4)^2-2*3*4=49-24=25
a^2+b^2=c^2?
c^2-a^2+b^2=0
second degree equation
c=(sqrt(4a^2+4b^2))/2
c=(sqrt(4*3^2+4*4^2)/2=5
(y+1)^2=Y^2+2y+1
if 2y+1=x^2
y+1= z
z^2=y^2+x^2
y=x^2-1/2
it seems that there is no demonstration for c^2
(y+1)^2-Y^2=2y+1
z^2-y^2=2y+1
Nothing says that 2y+1 is a square number.Except by digital applications.Otherwise we must go back to Greek mathematicians.