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November 12th, 2017, 09:33 PM   #1
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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.
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November 12th, 2017, 10:33 PM   #2
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what does $(a,b,c)=1$ mean?
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November 13th, 2017, 03:27 AM   #3
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It means all a and b and c are relatively prime to one another.
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November 13th, 2017, 04:15 AM   #4
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There are several proofs that show there is an infinite number of primitive Pythagorean triplets.
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November 13th, 2017, 07:18 AM   #5
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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.
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November 13th, 2017, 08:37 PM   #6
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Gave him the answer.

$\displaystyle x^2+y^2=z^4$

$\displaystyle x=2ps$

$\displaystyle y=p^2−s^2$

$\displaystyle z^2=p^2+s^2$

$\displaystyle z=k^2+t^2$

$\displaystyle p=2kt$

$\displaystyle s=k^2−t^2$

But why then is he not satisfied...
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November 13th, 2017, 09:50 PM   #7
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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.
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November 14th, 2017, 04:44 AM   #8
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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.
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November 15th, 2017, 12:00 AM   #9
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OOO, ok that makes alot more sense, thank you !
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