I don't get what to do with this one

[deft trellis]

Because 253x + 256y = 1

How will that have infinite solutions

[coarse elm]

it has infinite solutions, but you are looking for a specific one in the integers

[deft trellis]

The first one is just a guide to make it easier to plug in

[deft trellis]

For example we try x + y = 0

Then 3y = 1

But then y is not integer

So we try 1 now

Every 3rd number for a small value of x yields an integer for y

In the form 1(mod 3)

Which essentially means it has a remainder of 1 when divided by 3

So -5,-2,1,4,7,10

So on

Since it is +3y

Every third value would work

We want small values of x

But we also have to have integer numbers for y

So trying the first formula we see that if x + y = 1

We have 253 + 3y

That has to be equal to 1

And y is an integer

For that case

Try the other values

Well every 3rd number is divisible by 3

That adds up to a multiple of 3

0 is a multiple also

But in this case we want to subtract and get 1

We are subtracting an integer multiple of 3 since y is an integer and it is multiplied by 3

So 253(x + y)

Also has to leave a remainder of 1 when divided by 3

For example 3(5) + 1 - 3y

Suppose y = 1

3(5) + 1 - 3(1) =

3(5) - 3(1) + 1

= 3(4) + 1

So we can see 1 works for this example

But since every 3rd term is divisible by 3

We have that 1-3,1-6,1+3,1+6

Those number terms all create integer solutions for y

Which is what we need

When x + y is the least possible absolute value