$ 20n+2|2003n+2002\to 20n+2|3n+1802\to 10n+1|30n+18020\to 10n+1|18017=43*419$ contradition.

Can you explain how those simplifications work?

Euclidean Algorithm

So there is no solution in pozitive integers.

I just made a table on my calculator; there are indeed no positive integer solutions.

Making a table on one's calculator is unnecessary - I also assume it was done to check that $10n + 1 \mid 18017 = 43\cdot 419$ has no solution, since $20n + 2 \mid 2003n + 2002$ cannot be checked on a calculator. Replay. If $20n + 2$ divides $2003n + 2002$, then it also divides $20(2003n + 2002) - 2003(20n+2) = 36034 = 2\cdot 43\cdot 419$. Thus we need $20n + 2 \mid 2\cdot 43\cdot 419$, or $10n + 1 \mid 43\cdot 419$. Thus $10n + 1$ must be equal to one of $1, 43, 419$ or $43\cdot 419 = 18017$, and none is possible for a positive integer $n$.

balaylay wrote: Can you explain how those simplifications work? Through the Euclidean Algorithm, which can be found in the Introductory to Number Theory book.

For the division to occur $n$ must be even say $n=2k$ ,$k\ge 1$ $20m+1|100(20m+1)+3n+901$ $\implies 20m+1|3n+9$ bcoz if $a|b+c$ and $a|b$,$\implies a|c$ But the fraction must be an integer if it divides And $\frac{3m+9}{20m+1}$ does not an integer for $m=1,2,3,4$ . $\frac{3m+9}{20m+1}\ge 5$ $\implies m\le 10$ And checking cases for $m=1,2,...10$ There is no solution.

$20n+2\textnormal{ }|\textnormal{ }2003n+2002\textnormal{ }\Leftrightarrow\textnormal{ } 20n+2\textnormal{ }|\textnormal{ }2003n+2002 - 100(20n+2)\textnormal{ }\\ \Leftrightarrow20n+2\textnormal{ }|\textnormal{ }3n+1802\\\Rightarrow 20n+2\textnormal{ }|\textnormal{ }20(3n+1802)=60n+36040\\\Leftrightarrow 20n+2\textnormal{ }|\textnormal{ }60n+36040-3(20n+2)=36034=2*43*419$ Checking all factors of $36034$, we see that none can be written in the form $20n+2$ for positive $n$ (For nonegative $n$, we see that $0$ works). Hence, there are no solutions.

it is better to use euclidean algorithm as follows 2003n+2002=(100)(20n+2)+3n+1802 hence on dividing 2003n+2002 by 20n+2 we get remainder as 3n+1802 which on checking is not congruent modulo 20n+2 note that remainder must be lesser than divisor or quotient we must verify for n<=105

OG! $ 20n+2| 2003n + 2002.$ $\Longrightarrow$ $ 20n+2| (2003n + 2002).20- (20n+2).2003= 36034$ $\Longrightarrow$ $ 10n+1|18017$. However, the only positive divisors of 18017 are 1, 43, 419, 18017 none of which when equated to $10n+1$ gives $n$ a positive integer, Thus, there are no solutions.

paragdey01 wrote: For the division to occur $n$ must be even say $n=2k$ ,$k\ge 1$ $20m+1|100(20m+1)+3n+901$ $\implies 20m+1|3n+9$ bcoz if $a|b+c$ and $a|b$,$\implies a|c$ But the fraction must be an integer if it divides And $\frac{3m+9}{20m+1}$ does not an integer for $m=1,2,3,4$ . $\frac{3m+9}{20m+1}\ge 5$ $\implies m\le 10$ And checking cases for $m=1,2,...10$ There is no solution. Would you explain me how did you get $20m+1|100(20m+1)+3n+901$ and $20m+1|3n+9?$