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Vassilis Parassidis's user avatar
Vassilis Parassidis's user avatar
Vassilis Parassidis's user avatar
Vassilis Parassidis
  • Member for 13 years, 2 months
  • Last seen more than a month ago
  • Vancouver, BC, Canada
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Right angle parallelepipeds with the same greatest diagonal (the sides and greatest diagonal positive integers with g.c.d=1)
The second term should be $a^2=(q^2+q+1)^2$ and the third $b^2=(q^2+q)^2,(q+1)^2$, and so on. (one of the sides of the parallelepiped).
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Right angle parallelepipeds with the same greatest diagonal (the sides and greatest diagonal positive integers with g.c.d=1)
If we have $a^4=b^4+c^4+d^2$ then $a=(q^2+q+1)^2$ is the greatest diagonal. $b=(q^2+q)^2, (q+1)^2$, and so on. The other two sides derive from the above material.
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Solutions of the equation $X^4-DY^4=z^4$
If we set $k=a/2$ where $a$ an integer, then in the equation $4k^3+k=±m$ $m$ is always an integer and so the equation $X^4 –DY^4=Z^4$ is solvable.
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Solutions of the equation $X^4-DY^4=z^4$
The filled circle does mean multiplication. Thanks for your suggestions. I was hoping for a technique that yields infinite solutions, not just a lot.
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Transcendental numbers as infinite products of sides of squares
@Robert.Can you please express this transcendental number you are referring to in a closed form?
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Transcendental numbers as infinite products of sides of squares
@AaronMeyerowitz.The first product converges to 2/pi. This product is well known, the Vieta product. The second product converges to the square root of e.
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