# Y = 3 solution

## Other Factorizations

### Reformatting the input :

(1): "y3"   was replaced by   "y^3".

### Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

y^3-(y)=0

### Step  2  :

#### Pulling out like terms :

2.1     Pull out like factors :

y3 - y  =   y • (y2 - 1)

#### Trying to factor as a Difference of Squares :

2.2      Factoring:  y2 - 1

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2- AB + AB - B2 =
A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 1 is the square of 1
Check :  y2  is the square of  y1

Factorization is :       (y + 1)  •  (y - 1)

#### Equation at the end of step  2  :

y • (y + 1) • (y - 1) = 0

### Step  3  :

#### Theory - Roots of a product :

3.1    A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

#### Solving a Single Variable Equation :

3.2      Solve  :    y = 0

Solution is  y = 0

#### Solving a Single Variable Equation :

3.3      Solve  :    y+1 = 0

Subtract  1  from both sides of the equation :
y = -1

#### Solving a Single Variable Equation :

3.4      Solve  :    y-1 = 0

Add  1  to both sides of the equation :
y = 1

### Three solutions were found :

1.  y = 1
2.  y = -1
3.  y = 0
Sours: https://www.tiger-algebra.com/drill/y3=y/

## Linear equations with one unknown

### Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

0-y-(3)=0

### Step  1  :

#### Pulling out like terms :

1.1     Pull out like factors :

-y - 3  =   -1 • (y + 3)

### Step  2  :

#### Solving a Single Variable Equation :

2.1      Solve  :    -y-3 = 0

Add  3  to both sides of the equation :
-y = 3
Multiply both sides of the equation by (-1) :  y = -3

### One solution was found :

y = -3
Sours: https://www.tiger-algebra.com/drill/0-y=3/

## All integer solutions of $x^3-y^3=2020$.

Alternatively, if $p$ is a prime natural number that divides $x^2+xy+y^2$, then $$(2x+y)^2+3y^2=4(x^2+xy+y^2)\equiv 0\pmod{p}\,.$$ Thus, either $p$ divides both $x$ and $y$, or $\left(\dfrac{-3}{p}\right)=1$. Now, by quadratic reciprocity, $$1=\left(\dfrac{-3}{p}\right)=\left(\dfrac{p}{-3}\right)=\left(\dfrac{p}{3}\right)\,,$$ whence $p\equiv 1\pmod{3}$. Because $$(x-y)(x^2+xy+y^2)=x^3-y^3=2020=2^2\cdot 5\cdot 101$$ with $5\not\equiv 1\pmod{3}$ and $101\not\equiv 1\pmod{3}$, we conclude that $5$ and $101$ cannot divide $x^2+xy+y^2$. Thus, the only possible prime divisor of $x^2+xy+y^2$ is $2$, and if $2\not\equiv 1\pmod{3}$ is a factor of $x^2+xy+y^2$, we must have $2\mid x$ and $2\mid y$. Since $x^2+xy+y^2\geq 0$, this implies $$x^2+xy+y^2=1\text{ or }x^2+xy+y^2=4\,.$$ The only solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ to $x^2+xy+y^2=1$ are $$(x,y)=\pm (1,0),\pm(0,1),\pm(1,-1)\,.$$ The only solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ to $x^2+xy+y^2=4$ are $$(x,y)=\pm (2,0),\pm(0,2),\pm(2,-2)\,.$$ None of these solutions satisfies $x^3-y^3=2020$.

answered Jul 7 '20 at 16:02

BatominovskiBatominovski

$\endgroup$Sours: https://math.stackexchange.com/questions/3703969/all-integer-solutions-of-x3-y3-2020

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## Solution y = 3

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How to Solve a System of Equations Using Cramer's Rule: Step-by-Step Method

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