The equation $a^3 + b^3 = c^2$ has solution $(a, b, c) = (2,2,4)$.

Solution 1:

If you substitute $ (x z, y z, z^2) $ for $ (a,b,c) $ then you're looking for solutions to

$ x^3 z^3 + y^3 z^3 = z^4 $

Assuming $ z \neq 0 $, divide both sides by $ z^3 $

$ x^3 + y^3 = z $

Now choose anything for $ x $ and $ y $ and this gives you a $ z $ that satisfies the original equation. $ (2, 2, 4) $ corresponds to $ x = 1 $, $ y = 1 $, $ z = 1^3 + 1^3 = 2 $.

Solution 2:

Well you can and draw another formula. $$x^3+y^3=z^2$$

$$x=(b^2-a^2)(b^2+2ba-2a^2)c^2$$

$$y=(b^2-a^2)(2b^2-2ab-a^2)c^2$$

$$z=3(b^2-a^2)^2(a^2-ab+b^2)c^3$$

The most interesting thing there is that the formula that led, like should not give mutually simple solutions, but after sokrasheniya on common divisor can be obtained and are relatively prime solutions. This means that the formula itself describes as relatively prime so no. Coprime solutions - there are private solutions.

equations $$X^3+Y^3=Z^2$$ Can be expressed by integers $p,s,a,b,c$ . where the number of $c$ characterizes the degree of primitiveness.

$$X=(3a^2+4ab+b^2)(3a^2+b^2)c^2$$

$$Y=2b(a+b)(3a^2+b^2)c^2$$

$$Z=3(a+b)^2(3a^2+b^2)^2c^3$$

And more.

$$X=2b(b-a)(3a^2+b^2)c^2$$

$$Y=2b(b+a)(3a^2+b^2)c^2$$

$$Z=4b^2(3a^2+b^2)^2c^3$$

If we decide to factor $$X^3+Y^3=qZ^2$$

For a compact notation we replace :

$$a=s(2p-s)$$

$$b=p^2-s^2$$

$$t=p^2-ps+s^2$$

then:

$$X=qb(a+b)c^2$$

$$Y=qa(a+b)c^2$$

$$Z=qt(a+b)^2c^3$$

And the most beautiful solution. If we use the solutions of Pell's equation: $p^2-3a^2s^2=1$

by the way $a$ May appear as a factor in the decision and. Then the solutions are of the form::

$$X=qa(2p-3as)sc^2$$

$$Y=q(p-2as)pc^2$$

$$Z=q(p^2-3aps+3a^2s^2)c^3$$

If we change the sign : $Y^3-X^3=qZ^2$

Then the solutions are of the form:

$$X=qa(2p+3as)sc^2$$

$$Y=q(p+2as)pc^2$$

$$Z=q(p^2+3aps+3a^2s^2)c^3$$

Another solution of the equation: $X^3+Y^3=qZ^2$

$p,s$ - integers asked us.

To facilitate the calculations we make the change. $a,b,c$

If the ratio is as follows : $q=3t^2+1$

$$b=2q(q+2\mp{6t})p^2+6q(t\mp1)ps+(q-1\mp{3t})s^2$$

$$c=6q(q-2(1\pm{t}))p^2+6q(t\mp1)ps+3(1\mp{t})s^2$$

$$a=12q(1\mp{t})p^2+6(4t\mp{q})ps+3(1\mp{t})s^2$$

If the ratio is as follows: $q=t^2+3$

$$b=3(q-1)(1\pm{t})s^2+2(3\pm{(q-1)t})ps+(1\pm{t})p^2$$

$$c=3(6-(q-1)(q-3\mp{t}))s^2+6(1\pm{t})ps+(q-3\pm{t})p^2$$

$$a=3(6-(q-1)(1\mp{t}))s^2+6(1\pm{t})ps+(1\pm{t})p^2$$

Then the solutions are of the form:

$$X=2c(c-b)$$

$$Y=(c-3b)(c-b)$$

$$Z=3a(c-b)^2$$

Then the solutions are of the form:

$$X=2(c-b)c$$

$$Y=2(c+b)c$$

$$Z=4ac^2$$

If the ratio is as follows : $q=t^2+3$

$$c=6(q-4)(2\pm{t})p^2+4(6\pm{(q-4)t})ps+2(2\pm{t})s^2$$

$$b=3(24-(q-4)(q-3\mp{2t}))p^2+12(2\pm{t})ps+(q-3\pm{2t})s^2$$

$$a=3(24-(q-4)(4\mp{2t}))p^2+12(2\pm{t})ps+2(2\pm{t})s^2$$

If the ratio is as follows $q=3t^2+4$

$$c=q(-q+7(4\mp{3t}))p^2+6q(t\mp{1})ps+(q-4\mp{3t})s^2$$

$$b=3q(2q-7(1\pm{t}))p^2+6q(t\mp{1})ps+3(1\mp{t})s^2$$

$$a=21q(1\mp{t})p^2+6(7t\mp{q})ps+3(1\mp{t})s^2$$

Then the solutions are of the form :

$$X=2(3c-2b)c$$

$$Y=2(3c+2b)c$$

$$Z=12ac^2$$

Then the solutions are of the form :

$$X=(2b-c)b$$

$$Y=(2b+c)b$$

$$Z=2ab^2$$

Solution 3:

This is not what your exercise is expecting.

For the purpose to have a reference around, following is what we know about the equation:

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

According to $\S 14.3.1$ of Henri's Cohen's Number Theory, Vol II, Analytic and Modern Tools,
the integer solutions of above equation, subject to $\gcd(x,y) = 1$, can be grouped into three families. These families are disjoint, in the sense that any solution of the equation belongs to one and only one family.

The $s$ and $t$ below denote coprime integers satisfying corresponding congruences modulo $2$ and $3$. The three families/parametrizations, up to exchange of $x$ and $y$, are:

$$\begin{align} 1. & \begin{cases} x &= s(s+2t)(s^2-2ts+4t^2)\\ y &= -4t(s-t)(s^2+ts+t^2)\\ z &= \pm(s^2-2ts-2t^2)(s^4+2ts^3+6t^2s^2-4t^3s + 4t^4) \end{cases} & s \text{ is odd },\quad s \not\equiv t \pmod 3\\ \\ 2. & \begin{cases} x &= s^4-4ts^3-6t^2s^2-4t^3s + t^4\\ y &= 2(s^4+2ts^3+2t^3s+t^4)\\ z &= 3(s-t)(s+t)(s^4+2s^3t+6s^2t^2+2st^3+t^4) \end{cases} & s \not\equiv t \pmod 2,\quad s \not\equiv t \pmod 3\\ \\ 3. & \begin{cases} x &= -3s^4 + 6t^2s^2 + t^4\\ y &= 3s^4+6t^2s^2-t^4\\ z &= 6st(3s^4+t^4) \end{cases} & s \not\equiv t \pmod 2,\quad 3 \not| t \end{align} $$ For more details and derivation, please consult Cohen's book mentioned above.