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 November 12th, 2017, 09:33 PM #1 Newbie   Joined: Oct 2017 From: sweden Posts: 12 Thanks: 0 Another Diophantine problem How do I show that $a^2+b^2=c^4$ has infinitely many solutions with $(a,b,c)=1$? I mean I just am real stuck on solving problems like this, a clear proof would greatly help me understand.
 November 12th, 2017, 10:33 PM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,602 Thanks: 816 what does $(a,b,c)=1$ mean?
 November 13th, 2017, 03:27 AM #3 Newbie   Joined: Oct 2017 From: sweden Posts: 12 Thanks: 0 It means all a and b and c are relatively prime to one another.
 November 13th, 2017, 04:15 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,641 Thanks: 959 Math Focus: Elementary mathematics and beyond There are several proofs that show there is an infinite number of primitive Pythagorean triplets.
 November 13th, 2017, 07:18 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra For every odd pair $(s,t)$ with $s \gt t$ $$\left(\frac{s-t}2,\frac{s+t}2,st\right)$$ is a primitive Pythagorean triple (you'll need a proof of this). From there your result comes easily. Last edited by v8archie; November 13th, 2017 at 07:21 AM.
 November 13th, 2017, 08:37 PM #6 Member   Joined: Jul 2015 From: tbilisi Posts: 30 Thanks: 8 Gave him the answer. $\displaystyle x^2+y^2=z^4$ $\displaystyle x=2ps$ $\displaystyle y=p^2âˆ’s^2$ $\displaystyle z^2=p^2+s^2$ $\displaystyle z=k^2+t^2$ $\displaystyle p=2kt$ $\displaystyle s=k^2âˆ’t^2$ But why then is he not satisfied... Thanks from heinsbergrelatz
 November 13th, 2017, 09:50 PM #7 Newbie   Joined: Oct 2017 From: sweden Posts: 12 Thanks: 0 What do you mean you gave me the answer? I don't understand, also why are the variable k and t suddenly introduced? I don't really get the implication; could you explain further please? I get like why the idea on showing there are infinitely many Pythagorean triples, but I don't understand how we necessarily show that c a Pythagorean triple is of a square form. Technically, I don't understand two parts - showing there are infinite Pythagorean triples, and how c must be a square number. Last edited by skipjack; November 13th, 2017 at 10:18 PM.
 November 14th, 2017, 04:44 AM #8 Global Moderator   Joined: Dec 2006 Posts: 18,145 Thanks: 1418 What v8archie posted is incorrect. For example, s = 3, t = 1 leads to the triple (1, 2, 3), which isn't Pythagorean. However, if s and t are coprime odd perfect squares with s > t, $\displaystyle \left(\frac{s-t}2\right)^2 + \sqrt{st}^2 = \left(\frac{s+t}2\right)^2$. For example, (s, t) = (9, 1) leads to $\displaystyle 4^2 + 3^2 = 5^2$. Note that (s, t) = (49, 1) leads to $\displaystyle 24^2 + 7^2 = 25^2 = 5^4$. Regarding Individ's post, note that if $p > s$ and $(p, s, z)$ is a Pythagorean triple, so that $\displaystyle p^2 + s^2 = z^2\!$, $\displaystyle (2ps)^2 + \left(p^2 - s^2\right)^2 = \left(p^2 + s^2\right)^2 = z^4\!$, so $\displaystyle (2ps,\, p^2 - s^2,\, z)$ is a triple of the required type if coprime. Thanks from greg1313 and heinsbergrelatz
 November 15th, 2017, 12:00 AM #9 Newbie   Joined: Oct 2017 From: sweden Posts: 12 Thanks: 0 OOO, ok that makes alot more sense, thank you !

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