Can Two Numbers Satisfy These Exponential Conditions?

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In summary, the conversation discusses finding two different numbers, x and y, that satisfy the equations x^2 + y^2 = z^3 and x^3 + y^3 = z^2. It is mentioned that the cube and square on the right-hand side do not have to be of the same number. The speaker suggests setting x to 0, which leads to y^2 = z^3 and y^3 = w^2. By dividing the latter by the former, we find that y = w^2/z^3. The speaker then suggests letting w = z^2, which results in y = z and y^3 - y^2 = 0. This yields the pairs (0
  • #1
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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  • #2
The cube and the square on the RHS need not be of the same number.
 
  • #3
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
 
  • #4
Let's let $x=0$ so that we have:

\(\displaystyle y^2=z^3\)

\(\displaystyle y^3=w^2\)

Dividing the latter by the former, we have:

\(\displaystyle y=\frac{w^2}{z^3}\)

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

\(\displaystyle y=z\)

This implies:

\(\displaystyle y^3-y^2=y^2(y-1)=0\)

Because of the cyclical symmetry, this yields:

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

There may or may not be more pairs that work, but we have found at least one pair satisfying the problem. :)
 
  • #5
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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