## Other Factorizations

### Reformatting the input :

Changes made to your input should not affect the solution:

(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 by step solution :

### Step 1 :

### Step 2 :

#### Pulling out like terms :

2.1 Pull out like factors :

y^{3} - y = y • (y^{2} - 1)

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

2.2 Factoring: y^{2} - 1

Theory : A difference of two perfect squares, A^{2} - B^{2} can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A^{2} - AB + BA - B^{2} =

A^{2}- AB + AB - B^{2} =

A^{2} - B^{2}

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 : y^{2} is the square of y^{1}

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 :

- y = 1
- y = -1
- y = 0

## 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 by step solution :

### Step 1 :

#### Pulling out like terms :

1.1 Pull out like factors :

-y - 3 = -1 • (y + 3)

#### Equation at the end of step 1 :

### 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## 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

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Specialties. Just before the checkout, I met Natasha. She decided to replenish food supplies. Having argued a bit with her, I insisted that I was paying at the checkout.

## Solution y = 3

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