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November 12th, 2017, 08:33 PM  #1 
Newbie Joined: Oct 2017 From: sweden Posts: 19 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, 09:33 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,790 Thanks: 922 
what does $(a,b,c)=1$ mean?

November 13th, 2017, 02:27 AM  #3 
Newbie Joined: Oct 2017 From: sweden Posts: 19 Thanks: 0 
It means all a and b and c are relatively prime to one another.

November 13th, 2017, 03:15 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,765 Thanks: 1014 Math Focus: Elementary mathematics and beyond 
There are several proofs that show there is an infinite number of primitive Pythagorean triplets.

November 13th, 2017, 06:18 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,234 Thanks: 2412 Math Focus: Mainly analysis and algebra 
For every odd pair $(s,t)$ with $s \gt t$ $$\left(\frac{st}2,\frac{s+t}2,st\right)$$ 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 06:21 AM. 
November 13th, 2017, 07:37 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... 
November 13th, 2017, 08:50 PM  #7 
Newbie Joined: Oct 2017 From: sweden Posts: 19 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 09:18 PM. 
November 14th, 2017, 03:44 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,704 Thanks: 1530 
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, $\displaystyle \left(\frac{st}2\right)^2 + \sqrt{st}^2 = \left(\frac{s+t}2\right)^2$. 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. 
November 14th, 2017, 11:00 PM  #9 
Newbie Joined: Oct 2017 From: sweden Posts: 19 Thanks: 0 
OOO, ok that makes alot more sense, thank you !


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