Finding Integer Solutions to $1998a+1996b+1=ab$

[anemone]

Find all pairs $(a,\,b)$ of integers such that $1998a+1996b+1=ab$.

 

[mente oscura]

Find all pairs $(a,\,b)$ of integers such that $1998a+1996b+1=ab$.

Hello.

Spoiler

Observations:

1º) $$ab=odd$$

2º) $$\dfrac{1998a+1996b}{ab-1}=1$$

2.1) [tex]ab \equiv{3 } \mod(4)<br /> <br /> 2.2) $$If \ a \equiv{1 } \mod(4) \rightarrow{}b \equiv{3 } \mod(4)$$<br /> <br /> 2.3) $$If \ a \equiv{3 } \mod(4) \rightarrow{}b \equiv{1 } \mod(4)$$[tex]1996(a+b)+2a+1=ab[tex] <br /> $$1996=\dfrac{ab-2a-1}{a+b}=F(a,b)$$.(*)<br /> <br /> $$If \ a=3 \ and \ b=2 \rightarrow{}F(a,b)=0$$<br /> <br /> $$If \ a=7 \ and \ b=5 \rightarrow{}F(a,b)=2$$<br /> <br /> $$If \ a=11 \ and \ b=9 \rightarrow{}F(a,b)=4$$<br /> <br /> $$If \ a=15 \ and \ b=13 \rightarrow{}F(a,b)=6$$<br /> <br /> ...<br /> <br /> ...<br /> <br /> Succession "a": $$4n-1$$<br /> <br /> Succession "b": $$4n-3$$<br /> <br /> Succession "F(a,b)": $$2n-2$$<br /> <br /> Therefore, (*):<br /> <br /> $$F(a,b)=1996 \rightarrow{}n=999$$<br /> <br /> $$a=4*999-1=3995$$<br /> <br /> $$b=4*999-3=3993$$[/tex][/tex][/tex]

[tex][tex]$$<br /> Regards.$$[/tex][/tex]

 

[kaliprasad]

Spoiler

we have

$ab-1998a - 1996 b = 1$
add 1998 * 1996 on both sides
$ab - 1998a - 1996 b + 1998 * 1996 = (1998 * 1996) +1$
= $(1997+1)(1997-1) + 1$
= $1997 ^2$
or $(b-1998)(a- 1996) = 1997^2$
as 1997 is prime

so $(b- 1998, a - 1996) = ( 1997,1997)$
OR $(1 , 1997^2)$
OR $( 1997^2,1)$
OR $( -1997,-1997)$
OR $( -1, -1997^2)$
OR $( -1997^2,-1)$

rest is arithmetic

 

[Opalg]

Find all pairs $(a,\,b)$ of integers such that $1998a+1996b+1=ab$.

[sp]Write the equation as $1997(a+b) = (a+1)(b-1)$. Since $1997$ is prime, it follows that either $a+1$ or $b-1$ must be a multiple of $1997$.

Case 1. If $a+1 = 1997k$ then (after cancelling $1997$ from both sides) the equation becomes $1997k-1+b = k(b-1)$, from which $k = \dfrac{1-b}{1998-b} = 1 - \dfrac{1997}{1998-b}.$ Again using the fact that $1997$ is prime, we see that $1998-b$ must be $\pm1$ or $\pm 1997$. That gives four values for $b$, namely $1,\ 1997,\ 1999,\ 3995$. The corresponding values for $k$ are $0,\ -1996,\ 1998,\ 2$, with $a = -1,\ -3986013,\ 3990005,\ 3993$.

Case 2. If $b-1 = 1997k$ then the equation becomes $a + 1997k + 1 = k(a+1)$, from which $k = \dfrac{a+1}{a-1996} = 1 - \dfrac{1997}{1996-a}.$ Thus $a-1996$ must be $\pm1$ or $\pm 1997$. That gives four values for $a$, namely $-1,\ 1995,\ 1997,\ 3993$. The corresponding values for $k$ are again $0,\ -1996,\ 1998,\ 2$, with $b = 1,\ -3986011,\ 3990007,\ 3995$.

Two of those solutions appear in both sets, so there are six solutions altogether, namely $(a,b) = (-1,1),\ (3993,3995),\ (1995,-3986011),\ (1997,3990007),\ (-3986013,1997),\ (3990005,1999).$[/sp]

Edit. kaliprasad's method is neater, avoiding the duplicate solutions that I had.

 

[kaliprasad]

I had solved it at

http://mathhelpboards.com/challenge-questions-puzzles-28/find-integer-solutions-challenge-5837.html