integers (x,y) ---such that--- ( 1/x + 1/y ) = 1/22

[henh...@gmail.com]

------ pls wait 3+ days (Longer if you find it easy or trivial) before posting answers or hints.



For what integers (x,y)
is 1/22
expressible as ( 1/x + 1/y ) ?

[Edward Murphy]

On 8/29/2022 8:53 PM, henh...@gmail.com wrote:

> For what integers (x,y)
> is 1/22
> expressible as ( 1/x + 1/y ) ?

[spoiler space]































Both x and y must be non-zero. (If x is zero, then 1/x is infinitely
large, so 1/x + 1/y is also infinitely large. Similar for y.)

At least one of x, y must be positive. (If both are negative, then 1/x
and 1/y are also both negative, so 1/x + 1/y is also negative.)

Without loss of generality, assume that x > 0 and x >= y. (Each
resulting pair can be swapped.) Then consider:

1/22 = 1/x + 1/y
xy = 22y + 22x
(x - 22)y = 22x
y = 22x/(x - 22)

Now we can just look at x = 1, 2, 3, etc. and see which ones result in
y being an integer:

11, -22 -> 22/(22*11) - 11/(22*11)
18, -99 -> 22/(22*18) - 4/(22*18)
20, -220 -> 22/(22*20) - 2/(22*20)
21, -462 -> 22/(22*21) - 1/(22*21)

23, 506 -> 22/(22*23) + 1/(22*23)
24, 264 -> 22/(22*24) + 2/(22*24)
26, 143 -> 22/(22*26) + 4/(22*26)
33, 66 -> 22/(22*33) + 11/(22*33)
44, 44 -> 22/(22*44) + 22/(22*44)

Based on the right side, we can reverse-engineer a general pattern of

1/22 = 22/(22*x) + (x-22)/(22*x)

Now 22/(22*x) reduces to 1/x, leaving

1/y = (x-22)/(22*x)

y = (22*x)/(x-22)

i.e. we need 22*x to be divisible by x-22 (or, equivalently, by the
absolute value of x-22). So any prime factors of x-22 that aren't
covered by the prime factors of 22 (those being 2 and 11) must
instead be covered by the prime factors of x. But if something is a
prime factor of both x-22 and x, then it must also be a prime factor
of their difference (22), so again 2 or 11. And indeed this is what
we see play out above; the valid values of y (not including swapping
pairs around) are a subset of

(+1 or -1) * 2^a (0 <= a <= 2) * 11^b (0 <= b <= 2)

[henh...@gmail.com]

slightly diff. method, but i got the same 9 (=5+4) answer-pairs.




note that 22 is a product of 2 primes

i was wondering if we replace the 22 with another product of 2 primes
we'd get the same pattern of 9 (=5+4) answer-pairs.

[Edward Murphy]

On 9/5/2022 5:59 PM, henh...@gmail.com wrote:

> note that 22 is a product of 2 primes
>
> i was wondering if we replace the 22 with another product of 2 primes
> we'd get the same pattern of 9 (=5+4) answer-pairs.

I think we would.

Let z = x - 22, then

y = 22(z + 22)/z
= 22z/z + (22^2)/z
= 22 + (22^2)/z

So the potential solutions are those where z is a factor of
22^2 = 2^2 * 11^2
or else the additive inverse of such a factor.

Similarly, if we replace 22 with the product of any two distinct
primes p and q, then x - pq must be of the form
(-1)^a * p^b * q^c
where a is 0 or 1, b is 0 or 1 or 2, and c is 0 or 1 or 2.

Going back to 22 as a specific example, the possible values of z are
1, 2, 4, -1, -2, -4,
11, 22, 44, -11, -22, -44,
121, 242, 484, -121, -242, -484

Arranging these in ascending order and translating back to x, we get:

(z -> x, y)

z <= -22 leads to x <= 0 (invalid)

-11 -> 11, -22
-4 -> 18, -99
-2 -> 20, -220
-1 -> 21, -462

+1 -> 23, 506
+2 -> 24, 264
+4 -> 26, 143
+11 -> 33, 66
+22 -> 44, 44

z > +22 leads to x > y (swap of one of the above, or invalid)

In general, I think the valid answers always end up being the
following (without loss of generality, let p < q):

y = pqx/(x - pq)

z = x - pq

(z < -pq leads to x < 0, invalid. -pq, -ppq, -qq, -pqq, and -ppqq
all fall into this category.)

-q -> q, -pq
-p^2
-p
-1

+1
+p
+p^2
+q
+pq -> 2pq, 2pq

(z > +pq is a swap of one of the others, or invalid. ppq, qq, pqq, and
ppqq all fall into this category.)

The only part of the ascending sequence that varies is whether p^2 or q
is larger (and thus whether -p^2 or -q is smaller). Either way, pq is
still smaller than either p^2 or q (and -pq is smaller than either -p^2
or -q).

[Jonathan Dushoff]

More generally, if we can factor n = Product(p_i^a_i), then the number of decompositions is Product(2a_i+1).

We make two fractions as follows:

For each prime factor p_i, we put up to a_i factors in our first fraction OR in our second fraction, OR just leave it out (2a_i+1 possibilities). To avoid duplication, we agree to subtract fractions if the first one is larger, and add them if it's not. Since the product of our denominators divides n, the sum (or difference) can be written as x/n, and we can divide our expression by x to get an answer.

We can map any solution back to the original set of fractions by simply removing any common factors between the two denominators.

Jonathan