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# Non-linear Diophantine equations

A non- linear Diophantine equation is every Diophantine equation which is not linear. For instance, the equation $x^2 + 3y^3 = 35$ is a non-linear Diophantine equation.

There is no universal method for solving non-linear Diophantine equations, however, there are a number of “methods” that can help us in solving some special types of non-linear Diophantine equations. Through examples, we will show some of these methods. In fact, we can say that we are going to use some simple tricks that will help us in solving such equations.

Example 1. Solve the following equation in the set of integers:

$$xy + 5 y = 11.$$

Solution.

Dismantle the left side of the equation to the factors:

$$y(x + 5) = 11.$$

Therefore, the product of integer expressions $y$ and $x+5$ is equal to $11$, and that is only possible in the following cases:

a) $$y = 1, x+5= 11$$

b) $$y = 11, x+5 = 1$$

c) $$y = -1, x+5 = -11$$

d) $$y = -11, x+5 =- 1$$

This means that solutions of the given equation are the ordered pairs of numbers $(x, y)$:

$$(6, 1), (-4, 11), (-16, -1), (-6, -11).$$

Example 2. Solve the following equation in the set of integers:

$$xy + x – 6y – 9 = 0.$$

Solution.

By acting as in the previous example, we rewrite the equation in the following way:

$$xy + x – 6y – 9 =0$$

$$x( y+1) – 3(y +1) – 6 = 0$$

$$(y +1) (x -3) = 6$$

Since $x, y \in \mathbb{Z}$, it follows $x-3, y+1 \in \mathbb{Z}$. Therefore, we distinguish the following cases:

1.) $$x – 3 = 6, y + 1 = 1$$

2.) $$x -3 = 3, y +1 = 2$$

3.)  $$x-3 = 2, y+1 = 3$$

4.) $$x-3 = 1, y+1 = 6$$

5.) $$x-3 = 6, y+1 = -1$$

6.) $$x-3 = -3, y+1 = -2$$

7.) $$x-3 = -2, y+1 = -3$$

8.) $$x -3 = -1, y +1 = -6$$

The solutions are the ordered pairs $(x, y)$: $( 9, 0), (6, 1), (5, 2), (4, 5), (-3, -2), (0, -3), (1, -4), (3, -7)$.

Example 3.  Solve the following equation in the set of integers:

$$\frac{1}{x} + \frac{1}{y} = \frac{1}{14}.$$

Solution.

In this case we are going to express one of the unknowns $x$ or $y$; let it be $y$. By multiplying the equation with $14xy$, we have:

$$14y + 14 x = xy$$

$$14 x = y( x -14)$$

$$y = \frac{14x}{x-14}$$

If we divide the numerator with the denominator, we will obtain the quotient $14$ and the remainder $196$. Therefore:

$y = 14 + \frac{196}{x-14}.$$Since y must be an integer, that \frac{196}{x-14} also must be an integer, and that will be for: x = 15, x = 16, x = 18, x = 21, x= 28, x = 63, x = 112, x = 210, x = 13, x = 12, x = 10, x = 7, x = -14, x = -35, x = -84, x = -182, and for these values, we obtain by the order: y = 210, y = 112, y = 63, y= 42, y = 28, y= 18, y= 16, y= 15, y = -182, y =-84, y = -35, y = -14, y = 7, y= 10, y = 12, y= 13. The resulting ordered pairs (x, y) are the solutions to the given equation. Example 4. Solve the following equation in the set of integers:$$x^2 + 5 y = 213657.$$Solution. As we know, the square of the integer x^2 always ends with the one of the digits 0,1, 4, 5, 6 or 9, and 5 y ends with the digit 0 or 5. Thus, the number x^2 + 5y ends with the one of the digits 0,1, 4, 5, 6 or 9. Since the number 213657 ends with the digit 7, it follows that the given equation has no integer solutions. Example 5. Solve the following equation in the set of integers:$$ x^2 + 4y = 123.$$Solution. Since 4y is an even number for each integer y, that from the given equation follows that x must be an odd number. Therefore, x has a form: x = 2a + 1, a \in \mathbb{Z}. Substituting it in the given equation, we have:$$(2a + 1)^2 + 4y = 123\Longleftrightarrow 4a^2 + 4a + 1 + 4y = 123\Longleftrightarrow 4a^2 + 4a + 4y = 122\Longleftrightarrow 2a ^2 + 2a + 2y = 61.$$The left side of the equation will always be an even number, for each a, y \in \mathbb{Z}, and the right side of the equation is an odd number. Therefore, the equation x^2 + 4y = 123 has no integer solutions. Example 6. Solve the following equation in the set of integers:$$(x^2 – y^2)^2 = 1 – 16 y.$$Solution. The right side of the equation is a non- negative number, because it is equal to the square of an integer, therefore, y \ge 0. We then conclude that the left side of the equation is not less than (2y – 1)^2. Now, from the inequality (2y – 1)^2 \le 1 + 16 y we get y \le 5. Therefore, the right side of the equation can be equal to 1, 17, 33, 49, 65 or 81. Since the right side of the equation must be a complete square, that we take into account only 1, 49 and 81 as complete squares. There are three possible cases: 1.)$$y = 0, (x^2)^2 = 1$$2.)$$y = 3, (x^2)^2 = 49$$3.)$$y = 5, (x^2)^2 = 81 $$From the first case we have:$$(x, y) = (-1, 0), (1, 0),$$from the second:$$(x, y) = (-4, 3) , (4, 3),$$and from the third:$$(x, y) = (-4, 5), (4, 5).$$The obtained ordered pairs (x, y) are the solutions to the given equation. Example 7. Solve the following equation in the set of integers:$$3^x – 2^y = 1.$$Solution. The original equation we can write in the form$$ 3^x – 1 = 2^y.$$By using the formula$$a^n – b^n =(a – b) \cdot (a^{n-1} \cdot b^0 + a^{n-2} \cdot b^1 + \ldots + a^1 \cdot b^{n-2} + a^0 \cdot b^{n-1})$$we have:$$ 3^x – 1 = 2^y \Longleftrightarrow (3 – 1) \cdot ( 3^{x-1} + 3^{x-2} + \ldots + 3 + 1) = 2^y\Longleftrightarrow 3^{x-1} + 3^{x-2} + \ldots + 3 + 1 = \frac{2^y}{2}\Longleftrightarrow 3^{x-1} + 3^{x-2} + \ldots + 3 + 1 = 2 ^{y -1}.$$Therefore, x can be an even number or 1. If x = 1, then 3^0 = 2^{y-1}, that is, y = 1. In this case, the solution to the given equation is (x,y ) = (1,1). If x is an even number, that is, x = 2 t, t \in \mathbb{N}, then 3^{2t} – 1 = 2^y. It follows$$(3^2 – 1) ( 3^{2t -2} + 3^{2t -3} + \ldots 3 ^2 + 1) = 2^y\Longrightarrow 3^{2t -2} + 3^{2t -3} + \ldots 3 ^2 + 1 = 2^{y-3}$$If t is an odd number, then 2^{y-3} is an odd number, and that is only possible for y = 3. In this case, the solution is the ordered pair (x, y) = (2, 3). If t is an even number, that is t = 2k, k \in \mathbb{N}, then x = 4k. Now we have:$$3^{4k} – 2^y = 1 \Longleftrightarrow 81^k – 1 = 2^y \Longleftrightarrow (81 -1 ) ( 81^{k -1} + 81^{k -2} \ldots 81 + 1) = 2^y\Longleftrightarrow 80 \cdot ( 81^{k -1} + 81^{k -2} \ldots 81 + 1) = 2^y .$$Since the left side of the equation is divisible by 5, and right side is not, then we conclude that in this case the equation has no integer solutions. Therefore, all integer solutions to the given equation are:$$(x , y) \in \{(1,1), (2, 3) \}.$$Example 8. Solve the following equation in the set of integers:$$x^2 y^2 – x^2 – 8y^2 = 2xy.$$Solution. The one obvious solution to the given equation is (x,y) = (0,0). By transforming the equation, we have:$$x^2 y^2 – x^2 – 8y^2 = 2xy \Longleftrightarrow x^2 y^2 – 7y^2 = x^2 + y^2 + 2xy\Longleftrightarrow y^2(x^2 – 7) = (x + y)^2.$$Since the right side of the equation is the square of an integer, that the left side must be the square of an integer. Therefore, let x^2 – 7 = t, for some t \in \mathbb{Z}. Now we have:$$x^2 – 7 = t^2 \Longleftrightarrow x^2 – t^2 = 7 \Longleftrightarrow (|x| – |t|) (|x | + | t |) = 7\Longleftrightarrow (|x| – |t|) = 1 \bigwedge (|x | + | t |) = 7. $$It follows:$$2 |x | = 8 \Longrightarrow | x | = 4 \Longrightarrow x = \pm 4.$$We distinguish two cases: 1.)$$x = 4, y^2 ( x^2 – 7) = (x + y)^2$$2.)$$x = -4, y^2 ( x^2 – 7) = (x + y)^2$$For x = 4, we have:$$y^2 ( x^2 – 7) = (x + y)^2 \Longleftrightarrow y^2 ( 16 – 7) = (4 + y)^2\Longleftrightarrow 9y^2 = (4 + y) ^2\Longleftrightarrow 3y = 4 +y \bigvee 3y = -4 – y.$$It follows, y = 2 or y = -1. In this case, solutions to the given equation are:$$(x, y) = (4, 2), (4, -1).$$For x = -4, we have:$$y^2 ( x^2 – 7) = (x + y)^2 \Longleftrightarrow y^2 ( 16 – 7) = ( y – 4)^2 \Longleftrightarrow 9y^2 = (y – 4)^2\Longleftrightarrow 3y = y – 4 \bigvee 3y = 4 – y.$$It follows y = -2 or y = 1. Therefore, in this case, solutions to the given equation are:$$(x, y) = ( -4, -2) , ( -4, 1).$\$