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Topic: PPT (a^2, b^2, c) (Read 823 times) 

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


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dudiobugtron
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Re: PPT (a^2, b^2, c)
« Reply #1 on: Dec 23^{rd}, 2014, 2:53pm » 
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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.


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towr
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Re: PPT (a^2, b^2, c)
« Reply #2 on: Dec 23^{rd}, 2014, 11:17pm » 
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From Euclid's formula we get that a^{2} = m^{2}  n^{2} b^{2} = 2mn c = m^{2} + n^{2} with m > n, m  n = odd and m,n coprime. For b to be square and m,n coprime, we need m,n = x^{2}, 2 y^{2} ( and x odd) or m,n = 2 x^{2}, y^{2} ( and y odd) So if there is such a triple, it needs to conform to either a^{2} = x^{4}  4 y^{4}, b^{2} = 4 x^{2} y^{2}, c = x^{4} + 4 y^{4} or a^{2} = 4 x^{4}  y^{4}, b^{2} = 4 x^{2} y^{2}, c = 4 x^{4} + y^{4} So, can anyone find a square 4 x^{4}  y^{4} ?

« Last Edit: Dec 23^{rd}, 2014, 11:19pm by towr » 
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Christine
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Re: PPT (a^2, b^2, c)
« Reply #3 on: Dec 24^{th}, 2014, 12:59am » 
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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


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towr
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Re: PPT (a^2, b^2, c)
« Reply #4 on: Dec 24^{th}, 2014, 8:58am » 
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There are more solutions that don't work: y=0, z=2 x^{2} 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 nontrivial solutions.


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towr
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Re: PPT (a^2, b^2, c)
« Reply #5 on: Dec 24^{th}, 2014, 12:08pm » 
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a^{2} also has to be odd Which means that modulo 8, it's 1 (because it's not 0 or 4) 4 x^{4}  y^{4} modulo 8 is 0, 3, 4 or 7, so it can't equal a^{2} x^{4}  4 y^{4} modulo 8 is 0, 1, 4 or 5, which can equal a^{2} modulo 8. So if there is a solution, we need to look for it here.


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rloginunix
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Re: PPT (a^2, b^2, c)
« Reply #6 on: Dec 26^{th}, 2014, 8:08pm » 
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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: x^{4}+ y^{4} = z^{2} (1) Given the problem statement let us assume that we have found three integers "a", "b", and "c" such that: a^{4}+ b^{4} = c^{2} (2) and "c" is the smallest possible solution integer and gcd(a, b, c) = 1. Rewrite the left hand side terms as: (a^{2})^{2} + (b^{2})^{2} = c^{2} (3) From towr's Reply #2 we must have: a^{2} = m^{2}  n^{2} (4) b^{2} = 2mn (5) c = m^{2} + n^{2} (6) m > n, m  n = 2 + 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: a^{2} + n^{2} = m^{2} (9) Again, from towr's Reply #2 for the above to be true we must have: a = p^{2}  q^{2} (10) n = 2pq (11) m = p^{2} + q^{2} (12) p > q, p  q = 2 + 1, gcd(p, q) = 1 (13) Here I have switched the expressions for "a" and "n" around. From (5), since "b^{2}" must be a perfect square, it follows that: m = r^{2} (14) n = 2s^{2} (15) gcd(r, s) = 1 (16) Now "n" is expressed twice: once in (11) and once in (15), hence: 2pq = 2s^{2} (17) pq = s^{2} (18) Again, since "s^{2}" must be a perfect square: p = u^{2} (19) q = v^{2} (20) Square (19) and (20) separately and then sum them: u^{4} + v^{4} = p^{2} + q^{2} (21) Put (12) into (21): u^{4} + v^{4} = p^{2} + q^{2} = m (22) Put (14) into (22): u^{4} + v^{4} = p^{2} + q^{2} = m = r^{2} (23) From where it follows that: r 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 {a^{2}, b^{2}, c}. (may be someone can shorten the proof)


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rloginunix
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Re: PPT (a^2, b^2, c)
« Reply #7 on: Dec 28^{th}, 2014, 9:27pm » 
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If we change the plus sign to a minus in (1) above and shuffle the variables' names a bit: z^{4}  x^{4} = y^{2} (2.1) the question then becomes "does a primitive Pythagorean triple of the form {a^{2}, b, c^{2}} 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: (x^{2})^{2} + y^{2} = (z^{2})^{2} (2.2) x^{2} = 2mn (2.3) y = m^{2}  n^{2} (2.4) z^{2} = m^{2} + n^{2} (2.5) m < z (2.6) (2.5): m = p^{2}  q^{2} (2.7) n = 2pq (2.8) z = p^{2} + q^{2} (2.9) (2.3): m = r^{2} (2.10) n = 2s^{2} (2.11) (2.8), (2.11): pq = s^{2} (2.12) (2.12): p = u^{2} (2.13) q = v^{2} (2.14) (2.14), (2.13), (2.7), (2.10): u^{4}  v^{4} = p^{2}  q^{2} = m = r^{2} (2.15) (2.15): r m < z (2.16) which violates the Well Ordering Principle, "z" was assumed to be the smallest positive integer solution ...


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