**x**

^{2}+ y^{2}= z^{2}(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,

**z**is then divisible by 4 since:

^{2}**z**

^{2}= (2 * Z)^{2}= 4 * (Z^{2})(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

**x**cannot be divisible by

^{2}+ y^{2}**4**since:

**x**

= 4X

= 4[ X

^{2}+ y^{2}= (2 * X + 1)^{2}+ (2 * Y + 1)^{2}== 4X

^{2}+ 4X + 1 + 4Y^{2}+ 4Y + 1 == 4[ X

^{2}+ X + Y^{2}+ 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:

**x**

^{2}= z^{2}- y^{2}= (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)

z + y = (2v)

z - y = (2w)

(8) Which means that:

**(2u)**[From (5) and (7)]

^{2}= (2v)(2w)(9) Dividing both sides by

**4**gives us:

**u**

^{2}= 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

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 = p**

^{2}**w = q**

^{2}(13) And, we have our solution since:

**z = v + w = p**

^{2}+ q^{2}**y = v - w = p**

^{2}- q^{2}**x = 2u = 2pq**[Since

**u**means

^{2}= vw**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:

**(p**

^{2}+ q^{2})^{2}= (2pq)^{2}+ (p^{2}- q^{2})^{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[p**

^{2}+ q^{2}]**y = d[p**

^{2}- q^{2}]**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,

**6**.

^{2}+ 8^{2}= 36 + 64 = 100QED

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.

## 21 comments:

Larry -

I am definitely a math amateur, but have been thinking about this for a while and have a couple of questions.

I've been thinking about this for a while and have a simple way to identify whole number pythagorean triples. Consider any odd number x.

Then, y = (x2 - 1)/2

and z = (x2 + 1)/2

For example if x = 5 (x2 = 25)

then y = 12 and z = 13.

For x = 7 (x2 = 49)

y = 24 and z = 25.

This works because x2 = y + z

and y + 1 = z

So x2 + y2

= (y + z) + y2

= z + (y + y2)

= z + (y * z)

= z2

For example,

if x = 5

25 + (12 * 12) = (13 * 13)

(12 + 13) + (12 * 12) = (13 * 13)

13 + (13 * 12) = (13 * 13)

(13 * 13) = (13 * 13)

So, basically, x2 + y2 = z2 works with whole numbers because there is

a number, which when squared is the difference between n2 and (n-1)2.

Does this make sense?

Thanks for your post.

Yes, 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 Larry,

Could you please teach me how to solve the below Diophantine equation?

Find all solutions to the Diophantine equation x^2 + y^2 = 7z^2.

Thank you very much.

Jenny

Hi Jenny,

Are 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,

Thank you very much.

The question is right. It should be no integer solution. Thanks again for show me the way to solve it.

Jenny

Hi, Larry.

I 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,

I'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.

Sorry 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,

If 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.

You have answered my question.

best regards

cf

Hi Larry,

For 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,

For 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.

greetings!

can 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:

x2+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.

Sketch 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.

p 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

(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.

Post a Comment