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riddles >> medium >> PPT (a^2, b^2, c)
(Message started by: Christine on Dec 23rd, 2014, 1:04pm)

Title: PPT (a^2, b^2, c)
Post by Christine on Dec 23rd, 2014, 1:04pm
Is it possible to find a PPT of the form (a^2, b^2, c) ?
Why or why not?

Title: Re: PPT (a^2, b^2, c)
Post by dudiobugtron on Dec 23rd, 2014, 2:53pm
You don't need the primitive condition - if it's possible to find any pythagorean triple of that form, then you can find a primitive one.

Title: Re: PPT (a^2, b^2, c)
Post by towr on Dec 23rd, 2014, 11:17pm
From Euclid's formula we get that
a2 = m2 - n2
b2 = 2mn
c = m2 + n2
with m > n, m - n = odd and m,n coprime.

For b to be square and m,n coprime, we need
m,n = x2, 2 y2 ( and x odd) or
m,n = 2 x2, y2 ( and y odd)

So if there is such a triple, it needs to conform to either
a2 = x4 - 4 y4, b2 = 4 x2 y2, c = x4 + 4 y4 or
a2 = 4 x4 - y4, b2 = 4 x2 y2, c = 4 x4 + y4

So, can anyone find a square |4 x4 - y4| ?

Title: Re: PPT (a^2, b^2, c)
Post by Christine on Dec 24th, 2014, 12:59am
This is what I get:

4 x^4 - y^4 = z^2
(2 x^2 - y^2) (2 x^2 + y^2) = z^2

the only integer solution is x = y = z = 0

So, no PPT

Title: Re: PPT (a^2, b^2, c)
Post by towr on Dec 24th, 2014, 8:58am
There are more solutions that don't work: y=0, z=2 x2
Since z needn't be prime, I think it's too soon to declare there are no valid solutions.

For example if we had (2x - y)(2x + y) = z^2, then x=82, y=36, z=160 would be a solution. So these types of equations can have non-trivial solutions.

Title: Re: PPT (a^2, b^2, c)
Post by towr on Dec 24th, 2014, 12:08pm
a2 also has to be odd
Which means that modulo 8, it's 1 (because it's not 0 or 4)

4 x4 - y4 modulo 8 is 0, 3, 4 or 7, so it can't equal a2

x4 - 4 y4 modulo 8 is 0, 1, 4 or 5, which can equal a2 modulo 8. So if there is a solution, we need to look for it here.

Title: Re: PPT (a^2, b^2, c)
Post by rloginunix on Dec 26th, 2014, 8:08pm
Guys, correct me if I am wrong but it feels like you should be able to use Fermat's own method of infinite descend here. So all credits go to the master.

It is basically a proof by contradiction - we assume that a smallest (positive) integer solution exists, massage it some, and then arrive at a contradiction that yet another even smaller integer solution exists.

I have used so many intermediate variables that I have numbered all the statements to make finding a mistake easy.

We seek, in general, integer solutions for an equation:

x4+ y4 = z2     (1)


Given the problem statement let us assume that we have found three integers "a", "b", and "c" such that:

a4+ b4 = c2     (2)


and "c" is the smallest possible solution integer and gcd(a, b, c) = 1. Rewrite the left hand side terms as:

(a2)2 + (b2)2 = c2    (3)


From towr's Reply #2 we must have:

a2 = m2 - n2     (4)

b2 = 2mn     (5)

c = m2 + n2     (6)

m > n, m - n = 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/omega.gif + 1, gcd(m, n) = 1     (7)


This is adjusted for (3). From (6) it follows that:

m < c     (8)


Let us work with (4) since all its powers are 2's and we know how deal with them if we carry the "n" term over to the left side of the equal sign:

a2 + n2 = m2     (9)


Again, from towr's Reply #2 for the above to be true we must have:

a = p2 - q2     (10)

n = 2pq     (11)

m = p2 + q2     (12)

p > q, p - q = 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/gamma.gif + 1, gcd(p, q) = 1     (13)


Here I have switched the expressions for "a" and "n" around. From (5), since "b2" must be a perfect square, it follows that:

m = r2     (14)

n = 2s2     (15)

gcd(r, s) = 1     (16)


Now "n" is expressed twice: once in (11) and once in (15), hence:

2pq = 2s2     (17)

pq = s2     (18)


Again, since "s2" must be a perfect square:

p = u2     (19)

q = v2     (20)


Square (19) and (20) separately and then sum them:

u4 + v4 = p2 + q2     (21)


Put (12) into (21):

u4 + v4 = p2 + q2 = m     (22)


Put (14) into (22):

u4 + v4 = p2 + q2 = m = r2     (23)


From where it follows that:

r http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/leqslant.gif m < c    (24)


which contradicts the assumption that "c" is the smallest integer solution. Hence, you can not have a primitive Pythagorean triple of the form {a2, b2, c}.


(may be someone can shorten the proof)

Title: Re: PPT (a^2, b^2, c)
Post by rloginunix on Dec 28th, 2014, 9:27pm
If we change the plus sign to a minus in (1) above and shuffle the variables' names a bit:

z4 - x4 = y2     (2.1)


the question then becomes "does a primitive Pythagorean triple of the form {a2, b, c2} exist"? Looks like the answer is negative. The proof is almost the carbon copy of the one above except that at the end instead of a sum we take the difference.

In a condensed form, using towr's Reply #2 formulas:

(x2)2 + y2 = (z2)2     (2.2)

x2 = 2mn     (2.3)

y = m2 - n2     (2.4)

z2 = m2 + n2     (2.5)

m < z     (2.6)

(2.5):
m = p2 - q2     (2.7)

n = 2pq     (2.8)

z = p2 + q2     (2.9)

(2.3):
m = r2     (2.10)

n = 2s2     (2.11)

(2.8), (2.11):
pq = s2     (2.12)

(2.12):
p = u2     (2.13)

q = v2     (2.14)

(2.14), (2.13), (2.7), (2.10):
u4 - v4 = p2 - q2 = m = r2     (2.15)

(2.15):
r http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/leqslant.gif m < z    (2.16)


which violates the Well Ordering Principle, "z" was assumed to be the smallest positive integer solution ...



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