Question

Solving a Diophantine equation?

63* `63*x+70*y+75*z=91`

Answer 1

`{ (x = 5k+2), (y = -12k-15h-8), (z = 7k+14h+7) :}`

for any integers `k` and `h`

Explanation:

Given:

`63x+70y+75z=91`

First note that all terms are divisible by `7` except `75z`.

So we require `z=7w` for some `w`.

Then:

`9x+10y+75w=13`

Note that `10y+75w` is divisible by `5`.

Hence we require `9x-13` to be divisible by `5`.

Hence `x=5k+2` for some `k`

Then:

`45k+18+10y+75w=13`

So:

`9k+2y+15w+1=0`

Since `2y+1` is odd, we require exactly one of `k` and `w` to be odd and the other even. We can express this by putting `w=k+2h+1`

Then:

`0 = 9k+2y+15(k+2h+1)+1 = 24k+2y+30h+16`

So:

`12k+y+15h+8 = 0`

Then note that `12k` and `15h` are both divisible by `3`, so we require `y = 3f-8` for some integer `f`.

Then:

`4k+f+5h = 0`

So for any integers `k` and `h`, we can put `f = -4k-5h`.

Then:

`y = 3f-8 = 3(-4k-5h-8) = -12k-15h-8`

and we have a solution of the original equation, with:

`{ (x = 5k+2), (y = -12k-15h-8), (z = 7k+14h+7) :}`

Answer 2

See below.

Explanation:

`63*x+70*y+75*z=91`

This equation can be solved using a technique inspired on the sieve of Eratosthenes.

1) `x + y + 7/63 y + z + 12/63 z = 91/63`

now calling `a = 7/63 y + 12/63 z - 91/63` we have

`color(red)(x+y+z + a=0)` and

`63 a= 7 y + 12 z- 91` and now

2) `9a = y + z + 5/7z - 91/7`

now calling `b = 5/7z - 91/7` we have

`color(red)(9a=y+z+b)` and now

3) `7b=5z-91 rArr b+2/5b=z-91/5`

and now calling `c = 2/5b+91/5` we have

`color(red)(z=b+c)` and now

4) `5c=2b+91 rArr 2c+1/2 c=b +91/2` and calling

`d = 1/2c-91/2` we have `2c-b+d=0` and

`color(red)(c = 2d+91)`

but `z = b+c rArr z = 2c+d+2d+91 = 7d+3 xx 91`

now `y = 9a-(z+b) = 9a-12d-5 xx 91` and finally

`x = -(y+z+a) = 5d-10a+2 xx 91`

Resuming

`color(red)({(x =5d-10a+2 xx 91 ),(y = 9a-12d-5 xx 91 ),(z =7d+3 xx 91 ):}`