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My try:

Subtracting the eqns:

$a(b-c) = 11$

$a=1,b-c=11$ OR $a=11,b-c=1$

Substituting these values back int the original eqn. does not give an integral answer.

Thus number of ordered pairs = $0$.

Is this correct? [I am not very confident about this solution.]

Please also tell a general way of approaching problems in which we have to find the number of integral ordered pairs given some equations.

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  • $\begingroup$ Why don't you like this solution ? $\endgroup$
    – Peter
    Feb 5 '14 at 17:37
  • $\begingroup$ @Peter I just said I am not confident about it[Not that I don't like it].Is this then correct? $\endgroup$ Feb 5 '14 at 17:38
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Your solution is correct. Noe that $a=1,b-c=11$ and $a=11, b-c=1$ both lead to $a+b=12+c$. Then $33=ac+bc=(12+c)c$ indeed has no solution. [There's an alomost-soution: $c=-1$ gives $-33$; so I wonder if maybe part of the problem statement may be miscopied]

There isn't really "a general way of approaching problems in which we have to find the number of integral ordered pairs given some equations", else Fermat's Last Theorem would have been settled long ago.

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Your solution is correct.

The only thing I would be a bit more careful about is that if two integers multiply to get $11$, it may be that they are $-1$ and $-11$. For instance, in $a(b-c) = 11$, we could have $b - c = -11$. But as it turns out this cannot work because the other factor, $a$, has to be positive.

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