Question

Solve the following equation in natural numbers : `x²+y²=1997(x-y)`?

Answer 1

`(x, y) = (170, 145)` or `(x, y) = (1817, 145)`

Explanation:

The following proof is based on that in the book "An Introduction to Diophantine Equations: A Problem-Based Approach" by Titu Andreescu, Dorin Andrica, Ion Cucurezeanu.

Given:

`x^2+y^2=1997(x-y)`

Let `a = (x+y)` and `b = (1997-x+y)`

Then:

`a^2+b^2 = (x+y)^2+(1997-x+y)^2`

`=x^2+2xy+y^2+1997^2+x^2+y^2-2(1997(x-y)+xy)`

`=x^2+2xy+y^2+1997^2+x^2+y^2-2(x^2+y^2+xy)`

`=1997^2`

Hence we find:

`{(0 < a = x+y < 1997), (0 < b = 1997-x+y < 1997) :}`

Since `1997` is prime, `a` and `b` have no common factor greater than `1`.

Hence there exist positive integers `m, n` with `m > n` and no common factor greater than `1` such that:

`{ (1997 = m^2+n^2), (a=2mn), (b=m^2-n^2) :} color(white)(XX)"or"color(white)(XX){ (1997 = m^2+n^2), (a=m^2-n^2), (b=2mn) :}`

Looking at `1997 = m^2+n^2` in mod `3` and mod `5` arithmetic, we find:

`2 -= 1997 = m^2+n^2` (mod `3`) hence `m -= +-1` and `n -= +-1` (mod `3`)

`2 -= 1997 = m^2+n^2` (mod `5`) hence `m -= +-1` and `n -= +-1` (mod `5`)

That means that the only possibilities for `m, n` modulo `15` are `1, 4, 11, 14`.

In addition note that:

`m^2 in (1997/2, 1997)`

Hence:

`m in (sqrt(1997/2), sqrt(1997)) ~~ (31.6, 44.7)`

So the only possibilities for `m` are `34, 41, 44`

We find:

`1997 - 34^2 = 841 = 29^2`

`1997 - 41^2 = 316` not a perfect square.

`1997 - 44^2 = 61` not a perfect square.

So `(m, n) = (34, 29)`

So:

`(a, b) = (2mn, m^2-n^2) = (1972, 315)`

or

`(a, b) = (m^2-n^2, 2mn) = (315, 1972)`

`color(white)()`
If `(a, b) = (1972, 315)` then:

`{(x+y=1972), (1997-x+y=315):}`

and hence:

`(x, y) = (1817, 145)`

`color(white)()`
If `(a, b) = (315, 1972)` then:

`{(x+y=315), (1997-x+y=1972):}`

and hence:

`(x, y) = (170, 145)`