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

The equation $a^3 + b^3 = c^2$ has solution $(a, b, c) = (2,2,4).$ Find 3 more solutions in positige integers. [hint. look for solutions of the form $(a,b,c) = (xz, yz, z^2)$

Attempt:

So I tried to use the hint in relation to the triple that they gave that worked. So I observed that the triple that worked $(2,2,4)$ could also be written as $$2(1,1,2)$$ So i attempted to generalize it to $$n(1,1,2)$$ I thought this may work because I am using the "origin" of where the triple $(2,2,4)$ came from. This did not work unfortunately. So I am asking what else could I comsider to devise some system to find these triples? Because I highly doubt they're asking me to just shoot in the dark and pick random numbers

• Why not substitute $(a,b,c) = (xz,yz,z^2)$ into $a^3 + b^3 = c^2$ and see what you get? You have $x^3z^3 + y^3z^3 = z^4$. If $z \ne 0$, this reduces to $x^3 + y^3 = z$ which means... – achille hui Jan 19 '16 at 22:11
• – Lucian Jan 19 '16 at 22:54

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 $x = 1$, $y = 1$, $z = 1^3 + 1^3 = 2$.

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 :

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

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

$$Z=2ab^2$$

Here is an infinite number of solutions:

For any positive integer $c$, $x = (1+c^{3})$, $y =c(1+c^{3})$, and $z = (1+c^{3})^2$.

Check:

$x^3+y^3 =(1+c^3)^3+(c(1+c^3))^3 =(1+c^3)^3(1+c^3) =(1+c^3)^4$ and $z^2 =((1+c^{3})^2)^2 =(1+c^{3})^4$.

This is gotten from the following, which is a re-creation of some algebra of mine from many years ago:

To find solutions to $x^n+y^n = z^m$ where $(n, m) = 1$.

Since $(n, m) = 1$, there are $a$ and $b$ such that $am-bn = 1$ or $am=bn + 1$.

If $x = u^b$, $y = v^b$, $z = w^a$, then $u^{bn}+v^{bn} =w^{am} =w^{bn+1}$.

Let $v = cu$. Then $u^{bn}+v^{bn} =u^{bn}+(cu)^{bn} =u^{bn}+c^{bn}u^{bn} =u^{bn}(1+c^{bn})$.

If $w = 1+c^{bn}$, we want $w^{bn+1} =w\,w^{bn} =wu^{bn}$ or $u = w$.

Therefore, a solution is $x = (1+c^{bn})^b$, $y = cu =c(1+c^{bn})^b$, and $z = (1+c^{bn})^a$.

If $n=3$ and $m=2$, we want $2a-3b = 1$. This is satisfied by $a=2, b=1$.

We get $x = (1+c^{3})$, $y =c(1+c^{3})$, and $z = (1+c^{3})^2$.

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.

Just because sometimes you can ignore the hint (ha ha), one can also obtain a family of solutions by taking an existing solution, and multiplying both $x$ and $y$ by $k^2$, and multiplying $z$ by $k^3$ for some $k \in \mathbb{N}, k \geq 2$.

For instance, if we take $k = 2$, then the above solution is transformed as

$$(2, 2, 4) \to (8, 8, 32)$$

and note that

$$8^3+8^3 = 512 + 512 = 1024 = 32^2$$