Many thanks. Thinking out aloud: 10x+12y =4 - why not simplify to 5x + 6y =2 And see concept of complementary solution ax+by = 0 and particular solution (x=-2, y= 2) .
If a and b are relatively prime positive integers, prove that the Diophantine equation ax - by = c has infinitely many solutions in the positive integers. [Hint: There exist integers xo and Yo such that axo + byo = c. For any integer t, which is larger than both I xo I / b and I Yo I /a, a positive solution of the given equation is x = xo + bt, y =-(yo - at).]
I usually solve this using modular arithmetic 10x+12y=4 5x+6y =2 6y=2-5x ( mod 5) y=2 (mod 5) y=5k +2 Sub in the equation 5x + 6(5k+2) =2 5x +30k +12 =2 5x=-30k-10 x=-6k-2 This is the same answer as tge video with replacing k with -k
what if you are given something like 63x - 23y = -7. The gcd (63, -23) = 1 and 1 | -7. but do you solve the equation 63x - 23y = 1 and then multiply by -7? I'm faced with this problem and confused.
Yes, solve for 63x - 23y = 1, it will gives value x and y both positive, then general solutions will be (x, y)= {(x0 - bn), (y0 - an). Then multiply by -7 to get the final solution.
His explanation covers this. 0 is a multiple of the gcd, so solutions will exist. In fact, they wouldn't be hard to find. You won't even need to go through the gcd route. For example, 10x + 12y = 0, then x = -6y/5. To make x an integer, choose y = 5k, where k is an integer, and then you get x = -6k. So all integer solutions have the form (x,y) = (-6k, 5k), for any integer k.
