A^2 + b^2 = c^3

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  • #1
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a^2 + b^2 = c^3
(a,b,c - integers, different from each other)
Is there a solution to this?
 
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Answers and Replies

  • #2
Zurtex
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A trivial solution would be:

a = -1
b = 0
c = 1
 
  • #3
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Zurtex said:
A trivial solution would be:

a = -1
b = 0
c = 1
Of course, for a,b,c you can have:
-5, 10, 5
-17, 68, 17
... and so on.
But lets add a condition |a|<>|b|<>|c|, or just
a, b, c - positive integers, different from each other.
Is there a solution, any idea?
 
  • #4
NateTG
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There's also
a=x^3,b=0,c=x^2
 
  • #5
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frax said:
Of course, for a,b,c you can have:
-5, 10, 5
-17, 68, 17
... and so on.
But lets add a condition |a|<>|b|<>|c|, or just
a, b, c - positive integers, different from each other.
Is there a solution, any idea?

I wrote a program in C++ for this and got the following results for
a,b,c upto 30

11, 2, 5
18, 26, 10

Of course u can put - signs for a and b.
 
  • #6
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For any prime congruent to 1 mod 4, it is the sum of two squares. For that matter so is the product of two such cases:

(a^2+b^2)(c^2+d^2) = (ac=+/-bd)^2+(ad-/+bc)^2.

This then can be carried to any degree. For example: 5=2^2+1^2 and fits the bill. We have:

5^3=5^2+10^2, and better: 2^2+11^2.
 
  • #7
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:rofl: :rofl: :rofl:
 
  • #8
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Let there be a pythagorian triplet x^2 + y^2 = z^2. Now multiply both sides with z^4,
x^2*z^4 + y^2*z^4 = (xz^2)^2 + yz^2)^2 = z^6 = (z^2)^3.
Find individual pythagorian triplets and you will find corresponding solutions by this method.
 
  • #9
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Let x^2 + y^2 = z. Now multiply both sides with z^2,
x^2*z^2 + y^2*z^2 = xz^2 + yz^2 = z^3
 

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