Can Two Numbers Satisfy These Exponential Conditions?

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Discussion Overview

The discussion revolves around finding two different numbers that satisfy specific exponential conditions involving sums of their squares and cubes. The mathematical framework includes equations where the sum of squares equals a cube and the sum of cubes equals a square.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant proposes the equations: x² + y² = z³ and x³ + y³ = z², questioning their correctness.
  • Another participant suggests that the cube and the square on the right-hand side need not be of the same number, indicating a potential for different variables.
  • A further reply modifies the equations to x² + y² = z³ and x³ + y³ = w², maintaining the idea of distinct variables.
  • A participant explores the case where x = 0, leading to the simplification of the equations to y² = z³ and y³ = w², and derives a relationship between y, w, and z.
  • This exploration results in the identification of pairs (0,0), (0,1), and (1,0) as potential solutions, while acknowledging that there may be more pairs that satisfy the conditions.
  • Another participant praises the mathematical skill demonstrated in transforming applications into equations, emphasizing its importance.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of the cube and square being derived from the same number, with some proposing distinct variables. The discussion remains unresolved regarding the completeness of the identified solutions and whether additional pairs exist.

Contextual Notes

The exploration includes assumptions about the values of x, y, z, and w, and the implications of setting x to zero. The discussion does not resolve whether the identified pairs are exhaustive or if other solutions may exist.

mathdad
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Find two different numbers such that the sum of their squares shall equal a cube, and the sum of their cubes equal a square.

Set up:

x^2 + y^2 = z^3
x^3 + y^3 = z^2

Is this correct?
 
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The cube and the square on the RHS need not be of the same number.
 
greg1313 said:
The cube and the square on the RHS need not be of the same number.
x^2 + y^2 = z^3
x^3 + y^3 = w^2
 
Let's let $x=0$ so that we have:

$$y^2=z^3$$

$$y^3=w^2$$

Dividing the latter by the former, we have:

$$y=\frac{w^2}{z^3}$$

Suppose we let $w=z^2$...

$$y=z$$

This implies:

$$y^3-y^2=y^2(y-1)=0$$

Because of the cyclical symmetry, this yields:

$$(x,y)\in\{(0,0),(0,1),(1,0)\}$$

There may or may not be more pairs that work, but we have found at least one pair satisfying the problem. :)
 
Very impressive reply. This is, in my opinion, the best math skill to master. The ability to transform applications to equations is uniquely important.
 

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