MHB Can All Linear Diophantine Equations Be Solved?

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SUMMARY

Linear Diophantine equations can be analyzed for integer solutions based on the greatest common divisor (gcd) of their coefficients. The discussion confirms that if \(a\) and \(b\) are relatively prime, the equation \(ax + by = N\) has integer solutions for any integer \(N\). Specifically, the equation \(70x + 42y = 1409\) has no integer solutions due to \(\gcd(70, 42) = 14\) not dividing the right-hand side. Conversely, \(70x + 42y = 1428\) and \(2016x + 4031y = 2014201520162017\) do have integer solutions as their respective gcds divide the right-hand sides.

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Decide which of the following equations are true or false. If false, explain/provide a counterexample.

(a) If $a, b \in \mathbb{Z}$ are relatively prime, then $ax+by = N$ has integer solutions for any integer $N$.
(b) The equation $70x+42y = 1409$ has integer solutions
(c) The equation $70x+42y = 1428$ has integer solutions
(d) The equation $2016x+4031y = 2014201520162017$ has integer solutions.

I think (a) is true since relativity prime means $\gcd(a, b) = 1$ and $1|N$.
 
Last edited:
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(b) $\gcd(70, 42) = 14$ which does not divide the RHS, so it has integer no solutions.
(c) $\gcd(70, 42) = 14$ which divides the RHS, so it has integer solutions.
(d) $\gcd(2016,4031) = 1$ which clearly divides the RHS so it has integer solutions.

Could someone please verify whether this is correct?
 
Last edited:
Guest said:
(b) $\gcd(70, 42) = 14$ which does not divide the RHS, so it has integer no solutions.
(c) $\gcd(70, 42) = 14$ which divides the RHS, so it has integer solutions.
(d) $\gcd(2016,4031) = 1$ which clearly divides the RHS so it has integer solutions.

Could someone please verify whether this is correct?

all including for (a) specified in 1st post correct
 
kaliprasad said:
all including for (a) specified in 1st post correct
Thanks. Is there a quicker way to do these questions, perhaps using something like modular arithmetic?
 
Guest said:
Thanks. Is there a quicker way to do these questions, perhaps using something like modular arithmetic?

(b) can be done quicker, for instance mod 2.
More specifically, the left hand side is even, while the right hand side is odd.
 

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