Proving an identity to have solutions over all the integers

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The discussion centers on proving that the equation 3x + 2y = 5 has infinitely many integer solutions. A participant demonstrates the method by showing that if x = 3 and y = -2 is a solution, then adding a multiple of integers to these values can yield additional solutions. The example provided is 3*(2a+1) + 2*(1-3a) = 5, indicating how to manipulate the variables to maintain equality. Another equation, 4x + 3y = 10, is similarly explored, confirming the approach's validity. The conversation highlights the concept of generating infinite solutions for linear Diophantine equations.
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Hello,

I was looking at some math problems and one kind caught my attention. The idea was to prove that let's say 3x+2y=5 has infinitely many solutions over the integers.

Can someone show me the procedure how a problem like this might be solved?
 
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One solution is x=3, y=-2. Can you find a way to add some number a to 3 and subtract another number b from -2 so that when you plug 3+a and -2-b into the equation the a term and the b term cancel out?
 
Thank you man, I perfectly understood how to solve this.

3x+2y=5 => 3*(2a+1)+2*(1-3a)=5

And then just to practice I solved another one

4x+3y=10 => 4*(3n+1)+3*(2-4n)=10

Are they correct?
 
That looks good
 
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