I started with $a^2 + b^2 = c^2$ Then I got a problem when I thought what about having Ann number or set of numbers which will satisfy $a^2 + b^2 = c^2$ And also $a^2 + b^2 = d^3$ Both the equation simultaneously for example $3^2 + 4^2 = 25$ But $25$ is not a cubic number I want to know that sum of two square numbers can be equal to a square and a cubic number for $n^6$? I am not asking about Pythagorean triplets I am asking which satisfies $d^3$?
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You should let $a, b, c, d > 0$ to exclude the trivial cases. – Toby Mak Jun 09 '19 at 12:27
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1What kind of numbers are $$a,b,c,d$$? – Dr. Sonnhard Graubner Jun 09 '19 at 12:27
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I want the numbers a , b , c , d – Ankit Kumar Jun 09 '19 at 12:30
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5Your question is already answered in Variation of Pythagorean triplets: $x^2+y^2 = z^3$. – Toby Mak Jun 09 '19 at 12:31
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1So $a,b,c,d$ are real numbers? But $1^2 + 1^2 = (\sqrt{2})^2$ and $1^2+1^2 = (\sqrt[3]{2})^3$. – Toby Mak Jun 09 '19 at 12:32
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Since $c^2=d^3$, $c$ has to be a cube, and $d$ has to be a square. Also, $d^3$ has to be of the form $4n$ or $4n+1$... My 2 cents. – Nicolas FRANCOIS Jun 09 '19 at 12:32
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Here is an example - you can find others by searching for triangles for which the hypotenuse happens to be a cube. $$ 117^2 + 44^2 = 125^2 = 25^3. $$
You can always manufacture one with a cheap trick: Whenever $$ a^2 + b^2 = c^2 $$ you know $$ (ac^2)^2 + (bc^2)^2 = c^6 $$ which is both a square and a cube. For the $3-4-5$ triangle that yields $$ 75^2 + 100^2 = 125^2 = 25^3 . $$
Ethan Bolker
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