Let $a,b,c,d\in \mathbb{Z}$. Solve $$a^3=a(b^2+c^2+d^2)+2bcd$$
I've tried everything but I haven't been able to find a general solution.
Note: We may assume $\gcd(a,b,c,d)=1$ because of homogeneity.
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1$\begingroup$ This equation has the form: $$a(a^2-b^2-c^2-d^2)=2bcd$$ This means that relatively simple solutions no. Some is necessarily a multiple of. $\endgroup$– individJan 19, 2015 at 18:08
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$\begingroup$ @individ: Thanks, if you manage to solve it please post your solution. (I know there is a nice solution.) $\endgroup$– user45220Jan 19, 2015 at 18:13
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$\begingroup$ Source of the problem???? $\endgroup$– Will JagyJan 19, 2015 at 18:22
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$\begingroup$ Source of the nice solution? $\endgroup$– Dietrich BurdeJan 19, 2015 at 18:43
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$\begingroup$ @WillJagy Sorry, I should have been more clear there, I meant "I know" as in "there has got to be". The equation is from Ptolemy's Theorem (geometry). In the pythagorean theorem the equivalent equation is $a^2=b^2+c^2$ and notice that multiplying by $a$ gives $a^3=a(b^2+c^2)$ which is SLIGHTLY similar to the one in the question. With Ptolemy's theorem the equation is the one given in the question. I thought that since the Pythagorean theorem has a nice characterization for the integer solutions, Ptolemy's theorem should have one too. $\endgroup$– user45220Jan 19, 2015 at 18:58
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2 Answers
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The solution of the equation:
$$a^3=a(b^2+c^2+d^2)+2bcd$$
If you use Pythagorean triple.
$$x^2+y^2=z^2$$
Then the formula for the solution of this equation can be written.
$$a=z(zp^2-2yps+zs^2)$$
$$b=y(zp^2-2yps+zs^2)$$
$$c=zx(s^2-p^2)$$
$$d=2x(zp-ys)s$$
$p,s$ - any integer asked us.
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$\begingroup$ Excellent, so it's related to pythagorean triangles? I can show you the geometry diagram of this equation if you want $\endgroup$ Jan 22, 2015 at 17:24
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$\begingroup$ @user45220 Why? I don't like the geometric approach. I prefer hard algebra. $\endgroup$– individJan 22, 2015 at 17:29
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$\begingroup$ Geometry is the original language of God $\endgroup$ Jan 22, 2015 at 18:16
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$\begingroup$ @user45220 No. The number is the measure of all! Even logical reasoning may be flawed. Everything can be questioned - not yet decided, and not made the experiment. $\endgroup$– individJan 22, 2015 at 18:28
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Below mentioned equation from above has another parametrization, but without the use of a pythagorean triple,
$a^3=a(b^2+c^2+d^2)+2bcd$
$a=(k^2-4k+4)$
$b=(2k^2-4k)$
$c=-(7k^2+4k-4)$
$d=(2k^2-4k)$
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$\begingroup$ Could you perhaps indicate how you obtained that parameterisation? Other people referring to this question may not find it as obvious :) $\endgroup$ Jul 7, 2017 at 4:54