Integer Solution to $x^3+y^2=z^2$

Question

What is the non-zero integer general solution to $x^3+y^2=z^2$ ? I guess it is already solved in some book or paper, in that case plz help me to find that.

Edit:

1. One solution is - $n^3 = [(n)(n+1)/2]^2 - [(n)(n-1)/2]^2$ Found in here.

2. $(x,y,z)=\left(abuv,\frac{ab(bu^3-av^3)}{2},\frac{ab(bu^3+av^3)}{2}\right)$ where $bu^3\equiv av^3\pmod{2}$

Answer 1

We have: $x^3+y^2 = z^2 \Longrightarrow x^3 = z^2-y^2$.

We are looking for non-trivial integer solutions. So, let's assume $x>0,y>0,z>0$. Thus, $z>y>0$. Let's write $z = y+n$.

This gives:

$$x^3 = (y+n)^2 - y^2 = 2ny + n^2$$

So, $x^3 \equiv n^2 \pmod{2n}$ implies there exists a solution in this form. Let's try it out:

$$n=1: x^3 = (y+1)^2-y^2 = 2y+1 \Longrightarrow x^3\equiv 1 \pmod{2}\text{ and }y = \dfrac{x^3-1}{2}$$

Examples:

$$3^3 = \left(\dfrac{3^3-1}{2}+1\right)^2 - \left(\dfrac{3^3-1}{2}\right)^2 = 14^2-13^2 \\ 5^3 = \left(\dfrac{5^3-1}{2}+1\right)^2 - \left(\dfrac{5^3-1}{2}\right)^2 = 63^2-62^2 \\ \vdots$$

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$$n=2: x^3 = (y+2)^2-y^2 = 4y+4 \Longrightarrow x^3\equiv 0 \pmod{4} \text{ and }y = \dfrac{x^3-4}{4}$$

Examples:

$$4^3 = \left(\dfrac{4^3-4}{4}+2\right)^2 - \left(\dfrac{4^3-4}{4}\right)^2 = 17^2-15^2 \\ 8^3 = \left(\dfrac{8^3-4}{4}+2\right)^2 - \left(\dfrac{8^3-4}{4}\right)^2 = 129^2-127^2 \\ \vdots$$

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$$n=3: x^3 = (y+3)^2-y^2 = 6y+9 \Longrightarrow x^3 \equiv 3 \pmod{6} \text{ and }y = \dfrac{x^3-9}{6}$$

Examples:

$$3^3 = \left(\dfrac{3^3-9}{6}+3\right)^2-\left(\dfrac{3^3-9}{6}\right)^2 = 6^2-3^2 \\ 9^3 = \left(\dfrac{9^3-9}{6}+3\right)^2-\left(\dfrac{9^3-9}{6}\right)^2 = 123^2-120^2 \\ \vdots$$

Note: $(2n)^2 \equiv 0 \pmod{4n}$ and $(2n-1)^2 \equiv n \pmod{2(2n-1)}$ for all positive integers $n$. This yields:

$$\begin{array}{c|c}n & \text{equivalence} \\ \hline 1 & x^3 \equiv 1\pmod{2} \\ 2 & x^3 \equiv 0\pmod{4} \\ 3 & x^3\equiv 3 \pmod{6} \\ 4 & x^3\equiv 0\pmod{8} \\ 5 & x^3\equiv 5 \pmod{10} \\ 6 & x^3\equiv 0 \pmod{12} \\ 7 & x^3\equiv 7 \pmod{14} \\ 8 & x^3\equiv 0 \pmod{16} \\ 9 & x^3\equiv 9 \pmod{18} \\ \vdots & \vdots\end{array}$$

Answer 2

Equation $(x^3+y^2=z^2)$ has parametric solution:

$x=(2k-1)$

$y=(k-1)(2k-1)$

$z=k(2k-1)$

For, $k=7$ we get:

$(x,y,z)=(13,78,91)$

Answer 3

There are a lot of solutions. For example, parameterize a set of such solutions.

We have modulo $7$ the only cubes are $0$ and $\pm1$ and besides $t^6=1$ for all $t$ non divisible by $7$. Take for example the cube $1$ so we have $$1\equiv(z-y)(z+y)\pmod7$$ We can choose for example $z+y=t^5$ and $z-y=t$ so we get $$z=\frac{t^5+t}{2}\hspace{15mm}y=\frac{t^5-t}{2}$$ This allow us to get the parameterization in integers $$\begin{cases}x=t^2\\y=\dfrac{t^5-t}{2}\\z=\dfrac{t^5+t}{2}\end{cases}$$in relation with the identity $$(t^2)^3+\left(\frac{t^5-t}{2}\right)^2=\left(\frac{t^5+t}{2}\right)^2$$

Answer 4

Let $(x,y,z)\in\mathbb{Z}^3$ be such that $x^3+y^2=z^2$. Then, $$x^3=z^2-y^2=(z-y)(z+y)\,.$$ Write $$z-y=s\,\prod_{i=1}^l\,p_i^{3t_i}\,\prod_{j=1}^m\,q_j^{3u_j+1}\,\prod_{k=1}^n\,r_k^{3v_k+2}\,,$$ where

• $s\in\{-1,+1\}$,
• $p_1,p_2,\ldots,p_l$, $q_1,q_2,\ldots,q_m$, and $r_1,r_2,\ldots,r_n$ are pairwise distinct prime natural numbers, and
• $t_1,t_2,\ldots,t_l$, $u_1,u_2,\ldots,u_m$, and $v_1,v_2,\ldots,v_k$ are nonnegative integers.

Take $$a:=\prod_{j=1}^m\,q_j\,,$$ $$b:=\prod_{k=1}^n\,r_k\,,$$ and $$c:=s\,\prod_{i=1}^l\,p_i^{t_i}\,\prod_{j=1}^m\,q_j^{u_j}\,\prod_{k=1}^n\,r_k^{v_k}\,.$$ Then, $z-y=ab^2c^3$ with $a$ and $b$ being squarefree positive integers, and $c$ being an integer. Because $x^3=(z-y)(z+y)$, it follows immediately that $$z+y=a^2bd^3\,,$$ for some integer $d$. Consequently, $$x=\sqrt[3]{(z-y)(z+y)}=abcd\,,\tag{*}$$ $$y=\frac{z+y}{2}-\frac{z-y}{2}=\frac{ab(ad^3-bc^3)}{2}\,,\tag{#}$$ and $$z=\frac{z+y}{2}+\frac{z-y}{2}=\frac{ab(ad^3+bc^3)}{2}\,.\tag{$\star$}$$ For $y$ and $z$ to be integers, we need that $$ab(c-d)\equiv ab(ad^3\pm bc^3)\equiv 0\pmod{2}\,.$$

Thus, a solution $(x,y,z)$ is uniquely determined by

• two squarefree positive integers $a$ and $b$ which are relatively prime, and
• two integers $c$ and $d$ such that $ab(c-d)$ is divisible by $2$

via the formulae (*), (#), and ($\star$) above. If we relax the first condition by allowing $a$ and $b$ to be arbitrary integers, then more than one quadruples $(a,b,c,d)$ may produce the same triple $(x,y,z)$.

Here are some infinite families presented by other users. The infinite family $$(x,y,z)=\left(N,\dfrac{N(N-1)}{2},\dfrac{N(N+1)}{2}\right)$$ for $N\in \mathbb{Z}$ corresponds to $(a,b,c,d)=(N,1,1,1)$. The infinite family $$(x,y,z)=\left(N^2,\dfrac{N(N^4-1)}{2},\dfrac{N(N^4+1)}{2}\right)$$ for $N\in \mathbb{Z}$ corresponds to $(a,b,c,d)=(N,1,1,N)$.