Let $x,y,z$ be some positive integers. Is it true that we cannot find any positive integer $n$ for which $$ \frac{(x+y+z)^2}{x^2+y^2+z^2}=1+\frac{2}{3n}\,\,? $$
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For this General formula. http://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/738527#738527 – individ Aug 27 '15 at 16:19
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Why do you think it is related to find the solutions of $ax^2+by^2=cz^2$? – Paolo Leonetti Aug 27 '15 at 16:23
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It is generally linked to the solution of the equation. $ax^2+by^2+jz^2=qxy+dxz+cyz$ – individ Aug 27 '15 at 16:28
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Then you can set equivalently $a=b=j$ e $q=d=c=3n$. Does it really help? – Paolo Leonetti Aug 27 '15 at 16:31
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Not properly recorded. – individ Aug 27 '15 at 16:37
1 Answers
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Assuming the result for smaller $x+y+z$, it is inductively true in the case that any of $x,y,z$ is divisible by 3; this answer is a work in progress.
Expand and simplify to obtain:
$$\frac{xy+yz+xz}{x^2+y^2+z^2} = \frac{1}{3n}$$
or equivalently
$$3n(xy+yz+zx) = x^2+y^2+z^2$$
But the only way a sum of three squares can be a multiple of 3 is if all of them are multiples of 3 or all of them are not multiples of 3.
Former case: writing $x=3i, y=3j, z=3k$ we obtain
$$3n(ij+jk+ki) = i^2+j^2+k^2$$
We are done by inductive hypothesis: given any solution we may make a strictly smaller solution, but we have assumed for induction that no smaller solution exists. Therefore the only solution can be $x=y=z=0$, which is outside the allowable range.
Patrick Stevens
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Everything written above is correct. Then, we can start with a triple $(x,y,z)$ for which $\sum x^2 / \sum xy$ is an integer multiple of $3$ and $3\nmid xyz$.. – Paolo Leonetti Aug 27 '15 at 16:40
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