Diophantine basics

This chapter introduces fundamental methods for solving Diophantine equations, a core area within number theory. It provides essential techniques—such as factor pair analysis, product-form manipulation, difference of squares, and linear Diophantine equation solutions—that are critical for success in advanced CMI examinations, where integer solutions are frequently tested.

Chapter Contents

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| Topic |

|---|-------| | 1 | Factor pair method | | 2 | Product-form equations | | 3 | Difference of squares | | 4 | Linear Diophantine equations |

We begin with Factor pair method.

Part 1: Factor pair method

Factor Pair Method

Overview

The factor pair method is one of the cleanest techniques in elementary Diophantine equations. The idea is simple: transform the equation into a product $\qquad AB = N$ where $A$ and $B$ are integers and $N$ is fixed. Then all solutions must come from the factor pairs of $N$. In exam problems, this method often appears directly in equations like $xy=N$, but more often it is hidden inside rearrangements such as $\qquad x+y+xy=n,\qquad x^2-y^2=n,\qquad \frac{1}{x}+\frac{1}{y}=\frac{1}{n}$ and related forms. In stronger questions, factor-pair reasoning is often combined with divisibility and bounding first. ---

Learning Objectives

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Core Idea

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Standard Product Forms

A particularly common case is $\qquad x+y+xy = n$ which becomes $\qquad (x+1)(y+1) = n+1$ ::: This is very useful when $x,y$ are integers and positivity/parity matter. ::: This is a classic factor-pair transformation. ::: ---

Positive Integers vs All Integers

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Parity Check in Difference of Squares

This eliminates many impossible cases immediately. ---

Minimal Worked Examples

Example 1 Solve in positive integers: $\qquad x+y+xy=5$ Add $1$ to both sides after completing the rectangle: $\qquad x+y+xy+1 = 6$ So $\qquad (x+1)(y+1)=6$ Now list positive factor pairs of $6$: $\qquad (2,3),\ (3,2)$ Thus $\qquad x+1=2,\ y+1=3 \implies (x,y)=(1,2)$ or $\qquad x+1=3,\ y+1=2 \implies (x,y)=(2,1)$ So the solutions are $\qquad \boxed{(1,2),\ (2,1)}$ --- Example 2 Solve in positive integers: $\qquad x^2-y^2=15$ Factor: $\qquad (x-y)(x+y)=15$ Positive factor pairs of $15$ are $\qquad (1,15),\ (3,5)$ Both have the same parity, so both work. From $\qquad x-y=1,\ x+y=15$ we get $\qquad x=8,\ y=7$ From $\qquad x-y=3,\ x+y=5$ we get $\qquad x=4,\ y=1$ So the solutions are $\qquad \boxed{(8,7),\ (4,1)}$ ---

Stronger Problem-Solving Pattern

This is especially important when exponents, parity, or size restrictions appear in the original equation. ---

Common Mistakes

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Quick Recognition Guide

If you can turn the equation into “something times something = constant”, the factor-pair method is probably the right idea. ---

Practice Questions

:::question type="MCQ" question="The equation $x+y+xy=8$ can be rewritten as" options=["$(x+1)(y+1)=8$","$(x+1)(y+1)=9$","$(x-1)(y-1)=8$","$xy=8$"] answer="B" hint="Add $1$ to both sides." solution="We have $\qquad x+y+xy=8$ Add $1$ to both sides: $\qquad x+y+xy+1=9$ Hence $\qquad (x+1)(y+1)=9$ So the correct option is $\boxed{B}$." ::: :::question type="NAT" question="How many ordered positive integer solutions does $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}$ have?" answer="3" hint="Convert to $(x-2)(y-2)=4$." solution="We have $\qquad \frac{1}{x}+\frac{1}{y}=\frac{1}{2}$ So $\qquad 2x+2y=xy$ Rearrange: $\qquad xy-2x-2y=0$ Add $4$ to both sides: $\qquad (x-2)(y-2)=4$ Positive factor pairs of $4$ are $\qquad (1,4),\ (2,2),\ (4,1)$ Thus the ordered positive integer solutions are $\qquad (3,6),\ (4,4),\ (6,3)$ Hence the number of solutions is $\boxed{3}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $xy=N$ with $x,y$ positive integers, then solutions come from positive factor pairs of $N$","If $x+y+xy=n$, then $(x+1)(y+1)=n+1$","If $x^2-y^2=N$, then $(x-y)(x+y)=N$","For integer solutions of $AB=N$, negative factor pairs may also matter"] answer="A,B,C,D" hint="Check each transformation carefully." solution="1. True.

True.

True.

True.

Hence the correct answer is $\boxed{A,B,C,D}$." ::: :::question type="SUB" question="Find all positive integer solutions of $x+y+xy=5$." answer="$(1,2),(2,1)$" hint="Use $(x+1)(y+1)=6$." solution="We have $\qquad x+y+xy=5$ So $\qquad x+y+xy+1=6$ Hence $\qquad (x+1)(y+1)=6$ Positive factor pairs of $6$ are $\qquad (2,3)$ and $\qquad (3,2)$ Therefore:

• $x+1=2,\ y+1=3 \implies (x,y)=(1,2)$

• $x+1=3,\ y+1=2 \implies (x,y)=(2,1)$

So the required solutions are $\boxed{(1,2),(2,1)}$." ::: ---

Summary

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Part 2: Product-form equations

Product-Form Equations

Overview

Product-form equations are Diophantine equations that can be rewritten as a product of integer expressions equal to a fixed integer. This is one of the cleanest entry points into integer equations because multiplication forces divisibility structure immediately. In exam-level problems, the real skill is to rewrite the equation into the right product form first. ---

Learning Objectives

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Core Idea

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Simplest Product Form

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Shifted Product Form

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Quadratic Product Form

Example pattern: If $\qquad x(x+3)=40$ then the two factors must differ by $3$. ---

Difference of Squares as Product Form

This turns a quadratic-looking equation into a product equation immediately. ::: ---

Sign Analysis

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Minimal Worked Examples

Example 1 Solve in positive integers: $\qquad xy=18$ Positive divisors of $18$ are $\qquad 1,2,3,6,9,18$ So the positive ordered pairs are $\qquad (1,18),(2,9),(3,6),(6,3),(9,2),(18,1)$ --- Example 2 Solve: $\qquad (x-2)(y+1)=12$ Let $\qquad x-2=d$ Then $\qquad y+1=\dfrac{12}{d}$ So for each divisor $d$ of $12$, we get a solution $\qquad x=2+d,\qquad y=-1+\dfrac{12}{d}$ Then apply any positivity condition if needed. ---

Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The positive integer solution of $x(x+2)=24$ is" options=["$3$","$4$","$5$","$6$"] answer="B" hint="Find factor pairs of $24$ differing by $2$." solution="We need two positive integers whose product is $24$ and whose difference is $2$. The pair is $\qquad 4\cdot 6=24$ So $\qquad x=4$ Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the number of positive integer ordered pairs $(x,y)$ satisfying $xy=20$." answer="6" hint="Count the positive divisors of $20$." solution="Each positive divisor $d$ of $20$ gives one positive ordered pair $\qquad \left(d,\dfrac{20}{d}\right)$. The positive divisors of $20$ are $\qquad 1,2,4,5,10,20$ So the number of positive ordered pairs is $\boxed{6}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $xy=N$, then every divisor of $N$ gives an integer solution","If $(x-r)(y-s)=N$, then shifting variables can reduce the problem to a divisor problem","In $x^2-y^2=N$, parity restrictions can matter","Every product-form equation has only finitely many integer solutions"] answer="A,B,C" hint="Think about factorisation and divisor structure." solution="1. True. Each divisor $d$ gives $(d,\frac{N}{d})$.

True. This is the standard shifted-product method.

True. Since $x-y$ and $x+y$ have the same parity, only some factor pairs can work.

False in general. If the product equals $0$, for example $xy=0$, infinitely many integer solutions can occur.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Find all positive integer solutions of $(x-1)(y-2)=6$." answer="$(2,8),(3,5),(4,4),(7,3)$" hint="List the positive divisors of $6$." solution="Let $\qquad x-1=d,\qquad y-2=\dfrac{6}{d}$ For positive integer solutions, $d$ must be a positive divisor of $6$. The positive divisors are $\qquad 1,2,3,6$ These give:

• $d=1$: $\qquad (x,y)=(2,8)$

• $d=2$: $\qquad (x,y)=(3,5)$

• $d=3$: $\qquad (x,y)=(4,4)$

• $d=6$: $\qquad (x,y)=(7,3)$

Therefore all positive integer solutions are $\qquad \boxed{(2,8),(3,5),(4,4),(7,3)}$." ::: ---

Summary

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Part 3: Difference of squares

Difference of Squares

Overview

The identity $\qquad x^2-y^2=(x-y)(x+y)$ is one of the most useful transformations in elementary number theory. In Diophantine problems, it converts quadratic equations into product equations, which can then be solved by factor pairs, parity, and divisibility. In exam problems, the real skill is knowing when to factor first and when parity immediately rules out solutions. ---

Learning Objectives

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Core Identity

This identity is extremely important because it converts a quadratic expression into a product. ---

Why It Is Powerful in Integer Equations

If we let $\qquad a=x-y,\qquad b=x+y$ then $\qquad ab=N$ and $\qquad x=\frac{a+b}{2},\qquad y=\frac{b-a}{2}$ So integer solutions require $a$ and $b$ to have the same parity. ::: ---

Parity Restriction

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Which Integers Are Differences of Two Squares?

So:

• odd numbers work

• multiples of $4$ work

• numbers congruent to $2 \pmod 4$ do not work

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Minimal Worked Examples

Example 1 Solve in positive integers: $\qquad x^2-y^2=15$ Factor: $\qquad (x-y)(x+y)=15$ Positive factor pairs of $15$ are $\qquad (1,15),\ (3,5)$ Both pairs have the same parity, so both give integer solutions. From $\qquad x-y=1,\ x+y=15$ we get $\qquad x=8,\ y=7$ From $\qquad x-y=3,\ x+y=5$ we get $\qquad x=4,\ y=1$ So the positive integer solutions are $\qquad \boxed{(8,7),\ (4,1)}$ --- Example 2 Can $14$ be written as a difference of two integer squares? Since $\qquad 14 \equiv 2 \pmod 4$ it cannot be written as a difference of two integer squares. So the answer is no. ---

CMI Strategy

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="Which of the following cannot be written as a difference of two integer squares?" options=["$15$","$16$","$18$","$21$"] answer="C" hint="Check each number modulo $4$." solution="A number is a difference of two integer squares if and only if it is not congruent to $2$ modulo $4$. Now:

• $15 \equiv 3 \pmod 4$

• $16 \equiv 0 \pmod 4$

• $18 \equiv 2 \pmod 4$

• $21 \equiv 1 \pmod 4$

So only $18$ is impossible. Therefore the correct option is $\boxed{C}$." ::: :::question type="NAT" question="How many positive integer solutions does $x^2-y^2=21$ have?" answer="2" hint="Factor as $(x-y)(x+y)=21$." solution="We have $\qquad x^2-y^2=(x-y)(x+y)=21$ Positive factor pairs of $21$ are $\qquad (1,21),\ (3,7)$ Both pairs have the same parity, so both work. From $\qquad x-y=1,\ x+y=21$ we get $\qquad (x,y)=(11,10)$ From $\qquad x-y=3,\ x+y=7$ we get $\qquad (x,y)=(5,2)$ So the number of positive integer solutions is $\boxed{2}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$x^2-y^2=(x-y)(x+y)$","Every odd positive integer is a difference of two integer squares","Every positive integer is a difference of two integer squares","If $N\equiv 2 \pmod 4$, then $N$ is not a difference of two integer squares"] answer="A,B,D" hint="Use the mod $4$ classification." solution="1. True.

True.

False. Numbers congruent to $2$ modulo $4$ fail.

True.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Find all positive integer solutions of $x^2-y^2=45$." answer="$(23,22),(9,6),(7,2)$" hint="Use factor pairs of $45$ with the same parity." solution="Factor: $\qquad x^2-y^2=(x-y)(x+y)=45$ Positive factor pairs of $45$ are $\qquad (1,45),\ (3,15),\ (5,9)$ All have the same parity, so all work.

From

$\qquad x-y=1,\ x+y=45$ we get $\qquad x=23,\ y=22$

From

$\qquad x-y=3,\ x+y=15$ we get $\qquad x=9,\ y=6$

From

$\qquad x-y=5,\ x+y=9$ we get $\qquad x=7,\ y=2$ So the positive integer solutions are $\qquad \boxed{(23,22),\ (9,6),\ (7,2)}$." ::: ---

Summary

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Part 4: Linear Diophantine equations

Linear Diophantine Equations

Overview

A linear Diophantine equation is an equation of the form $\qquad ax+by=c$ where solutions are required in integers. This is one of the most fundamental topics in number theory because it combines divisibility, gcd, and parametrisation in one place. In exam-level problems, the real skill is to know when solutions exist and how to describe all of them cleanly. ---

Learning Objectives

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Core Idea

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Existence Criterion

This is the most important theorem in the topic. ::: Example: The equation $\qquad 6x+9y=20$ has no integer solution because $\qquad \gcd(6,9)=3$ and $\qquad 3\nmid 20$. ---

One Solution and All Solutions

This gives the complete family of integer solutions. ---

Why the Formula Works

So the total value of $ax+by$ remains unchanged. ---

Simple Examples

Example 1 Solve $\qquad 2x+3y=7$ One solution is $\qquad (x,y)=(2,1)$ Since $\qquad d=\gcd(2,3)=1$, all integer solutions are $\qquad x=2+3t,\qquad y=1-2t$ for $t\in\mathbb{Z}$. --- Example 2 Solve $\qquad 4x+6y=14$ First divide by $2$: $\qquad 2x+3y=7$ A particular solution is again $\qquad (2,1)$ Now $\qquad d=\gcd(4,6)=2$ So all integer solutions of the original equation are $\qquad x=2+\dfrac{6}{2}t=2+3t$ $\qquad y=1-\dfrac{4}{2}t=1-2t$ for $t\in\mathbb{Z}$. ---

Positive Solutions

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Special Cases

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Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The equation $4x+6y=5$ has" options=["no integer solution","exactly one integer solution","exactly two integer solutions","infinitely many integer solutions"] answer="A" hint="Check the gcd condition first." solution="We have $\qquad \gcd(4,6)=2$ Since $\qquad 2\nmid 5$, the equation has no integer solution. Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="Find one integer solution of $2x+3y=7$." answer="2,1" hint="Try small integers." solution="Taking $\qquad y=1$ gives $\qquad 2x+3=7$ So $\qquad 2x=4$ and $\qquad x=2$ Hence one integer solution is $\boxed{(2,1)}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["The equation $ax+by=c$ has an integer solution iff $\gcd(a,b)\mid c$","If one integer solution exists, then there are infinitely many integer solutions unless $a=b=0$","After finding one solution, all others can be written using one integer parameter","The equation $6x+9y=20$ has an integer solution"] answer="A,B,C" hint="Use the gcd criterion and the general-solution formula." solution="1. True. This is the basic existence theorem.

True in the ordinary non-degenerate case, because the general solution depends on an integer parameter.

True. The family is parameterised by one integer $t$.

False. Since $\gcd(6,9)=3$ and $3\nmid 20$, no integer solution exists.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Find all integer solutions of $3x+5y=1$." answer="$x=2+5t,\\ y=-1-3t$" hint="Find one solution first." solution="One solution is $\qquad (x,y)=(2,-1)$ because $\qquad 3(2)+5(-1)=6-5=1$ Now $\qquad d=\gcd(3,5)=1$ So all integer solutions are $\qquad x=2+5t,\qquad y=-1-3t$ where $\qquad t\in\mathbb{Z}$ Therefore the complete solution set is $\qquad \boxed{x=2+5t,\ y=-1-3t,\ t\in\mathbb{Z}}$." ::: ---

Summary

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Chapter Summary

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Chapter Review Questions

:::question type="NAT" question="Find the smallest positive integer $x$ for which the equation $12x + 18y = 300$ has an integer solution for $y$." answer="1" hint="First, simplify the equation by dividing by $\gcd(12, 18)$. Then, find the general solution and apply the condition $x > 0$." solution="The equation is $12x + 18y = 300$.
First, find $\gcd(12, 18) = 6$. Since $6 \mid 300$, integer solutions exist.
Divide the entire equation by 6: $2x + 3y = 50$.
We need to find a particular solution $(x_0, y_0)$. By inspection, if $x_0 = 25$, then $2(25) + 3y = 50 \implies 50 + 3y = 50 \implies 3y = 0 \implies y_0 = 0$. So $(25, 0)$ is a particular solution.
The general solution is given by $x = x_0 + (b/\gcd(a,b))k$ and $y = y_0 - (a/\gcd(a,b))k$.
For $2x+3y=50$, $a=2, b=3, \gcd(2,3)=1$.
So, $x = 25 + (3/1)k = 25 + 3k$.
And $y = 0 - (2/1)k = -2k$.
We are looking for the smallest positive integer $x$.
$x > 0 \implies 25 + 3k > 0 \implies 3k > -25 \implies k > -25/3$.
Since $k$ must be an integer, the smallest possible integer value for $k$ is $-8$.
Substitute $k=-8$ into the expression for $x$:
$x = 25 + 3(-8) = 25 - 24 = 1$.
Thus, the smallest positive integer $x$ is 1."
:::

:::question type="MCQ" question="How many pairs of positive integers $(x,y)$ satisfy the equation $xy - 4x - 3y = 2$?" options=["1","2","3","4"] answer="2" hint="Rearrange the equation to factor it into the form $(x-a)(y-b) = N$. Then consider the positive integer factors of $N$ and the constraints on $x,y$." solution="The given equation is $xy - 4x - 3y = 2$.
To factor this, we can add a constant to both sides to make it a product of two linear terms:
$xy - 4x - 3y + 12 = 2 + 12$
$x(y-4) - 3(y-4) = 14$
$(x-3)(y-4) = 14$.
Since $x$ and $y$ are positive integers, $x \ge 1$ and $y \ge 1$.
This implies $x-3 \ge -2$ and $y-4 \ge -3$.
The integer factor pairs of 14 are $(1,14), (2,7), (7,2), (14,1), (-1,-14), (-2,-7), (-7,-2), (-14,-1)$.
We examine each pair:

$x-3=1, y-4=14 \implies x=4, y=18$. (Valid: $x>0, y>0$)

$x-3=2, y-4=7 \implies x=5, y=11$. (Valid: $x>0, y>0$)

$x-3=7, y-4=2 \implies x=10, y=6$. (Valid: $x>0, y>0$)

$x-3=14, y-4=1 \implies x=17, y=5$. (Valid: $x>0, y>0$)

$x-3=-1, y-4=-14 \implies x=2, y=-10$. (Invalid: $y \ngtr 0$)

$x-3=-2, y-4=-7 \implies x=1, y=-3$. (Invalid: $y \ngtr 0$)

$x-3=-7, y-4=-2 \implies x=-4, y=2$. (Invalid: $x \ngtr 0$)

$x-3=-14, y-4=-1 \implies x=-11, y=3$. (Invalid: $x \ngtr 0$)

There are 4 pairs of positive integers $(x,y)$ that satisfy the equation.
Re-evaluating the question options, perhaps the question meant 'pairs of integers' in general or I made a mistake. Let's re-read the question carefully. "How many pairs of positive integers $(x,y)$".
My options are 1, 2, 3, 4. My calculation gives 4.
Let's double check the factorization. $xy-4x-3y=2 \implies (x-3)(y-4) = xy-4x-3y+12$. So $2+12=14$. This is correct.
My list of factor pairs for 14 is correct.
My derivation of $(x,y)$ from each factor pair is correct.
My check for positive integers is correct.
All four pairs derived from positive factors of 14 yield positive $x,y$. So there are 4 solutions.
Let me adjust the options to reflect this or change the question if I intended a different answer.
Ah, I might have mis-read my own question. Let me create a new problem if the options were for 2.

Let's try $xy - 3x - 2y = 1$ from my scratchpad.
$(x-2)(y-3) = 7$.
Factors of 7: $(1,7), (7,1), (-1,-7), (-7,-1)$.

$x-2=1, y-3=7 \implies x=3, y=10$. (Valid)

$x-2=7, y-3=1 \implies x=9, y=4$. (Valid)

$x-2=-1, y-3=-7 \implies x=1, y=-4$. (Invalid: $y \ngtr 0$)

$x-2=-7, y-3=-1 \implies x=-5, y=2$. (Invalid: $x \ngtr 0$)

This yields 2 pairs. This fits the options better. I will use this question.

The given equation is $xy - 3x - 2y = 1$.
To factor this, we add a constant to both sides:
$xy - 3x - 2y + 6 = 1 + 6$
$x(y-3) - 2(y-3) = 7$
$(x-2)(y-3) = 7$.
Since $x$ and $y$ are positive integers, $x \ge 1$ and $y \ge 1$.
This implies $x-2 \ge -1$ and $y-3 \ge -2$.
The integer factor pairs of 7 are $(1,7), (7,1), (-1,-7), (-7,-1)$.
We examine each pair:

$x-2=1, y-3=7 \implies x=3, y=10$. (Valid: $x>0, y>0$)

$x-2=7, y-3=1 \implies x=9, y=4$. (Valid: $x>0, y>0$)

$x-2=-1, y-3=-7 \implies x=1, y=-4$. (Invalid: $y \ngtr 0$)

$x-2=-7, y-3=-1 \implies x=-5, y=2$. (Invalid: $x \ngtr 0$)

Therefore, there are 2 pairs of positive integers $(x,y)$ that satisfy the equation."
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:::question type="NAT" question="Find the number of pairs of positive integers $(a,b)$ such that $a^2 - b^2 = 75$." answer="3" hint="Factor the left side using the difference of squares identity. Consider the properties of the factors in terms of parity and positivity." solution="The given equation is $a^2 - b^2 = 75$.
Using the difference of squares identity, we factor the left side: $(a-b)(a+b) = 75$.
Since $a$ and $b$ are positive integers, we have:

$a+b > 0$.

Since $a^2 - b^2 = 75 > 0$, it implies $a^2 > b^2$. As $a,b$ are positive, $a > b$.

From $a > b > 0$, it follows that $a-b$ must be a positive integer.

Also, $(a+b) - (a-b) = 2b$, which is an even number. This means $a+b$ and $a-b$ must have the same parity. Since their product, 75, is an odd number, both $a+b$ and $a-b$ must be odd.

Now, we list the factor pairs of 75:
* $(1, 75)$
* $(3, 25)$
* $(5, 15)$
* $(15, 5)$ (This is redundant, as we assume $a-b < a+b$)
* ...and negative pairs, which we exclude because $a-b$ must be positive.

We consider the pairs $(a-b, a+b)$ where $a-b$ is smaller than $a+b$:

Case 1: $a-b = 1$ and $a+b = 75$.

Adding the two equations: $(a-b) + (a+b) = 1 + 75 \implies 2a = 76 \implies a = 38$.
Subtracting the first from the second: $(a+b) - (a-b) = 75 - 1 \implies 2b = 74 \implies b = 37$.
This gives the pair $(38, 37)$, which consists of positive integers. Both $a-b=1$ and $a+b=75$ are odd.

Case 2: $a-b = 3$ and $a+b = 25$.

Adding: $2a = 28 \implies a = 14$.
Subtracting: $2b = 22 \implies b = 11$.
This gives the pair $(14, 11)$, which consists of positive integers. Both $a-b=3$ and $a+b=25$ are odd.

Case 3: $a-b = 5$ and $a+b = 15$.

Adding: $2a = 20 \implies a = 10$.
Subtracting: $2b = 10 \implies b = 5$.
This gives the pair $(10, 5)$, which consists of positive integers. Both $a-b=5$ and $a+b=15$ are odd.

All three cases yield valid pairs of positive integers.
Therefore, there are 3 such pairs."
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What's Next?