MHB Can All Linear Diophantine Equations Be Solved?

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The discussion centers on the solvability of linear Diophantine equations. It is established that if \(a\) and \(b\) are relatively prime, then \(ax + by = N\) has integer solutions for any integer \(N\). For the specific equations, it is concluded that (a) is true, (b) is false due to the GCD not dividing the right-hand side, (c) is true as the GCD divides the RHS, and (d) is true since the GCD is 1. Additionally, participants suggest using modular arithmetic for quicker verification, particularly noting that equation (b) can be assessed by checking parity. The discussion emphasizes the importance of GCD in determining the existence of integer solutions.
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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 $1|N$.
 
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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?
 
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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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