x, y, 24?

If $xy+1 \equiv 0 \pmod{24},$ then $xy+1 \equiv 0 \pmod{8}$ and $xy+1 \equiv 0 \pmod{3}.$

If $xy+1 \equiv 0 \pmod{8},$ then one of these must be true:

$\begin{cases}\begin{aligned} (x,y) &\equiv (1,7) \pmod{8} \\ (x,y) &\equiv (7,1) \pmod{8} \\ (x,y) &\equiv (3,5) \pmod{8} \\ (x,y) &\equiv (5,3) \pmod{8} \end{aligned}\end{cases}$

In any of these cases, $x+y \equiv 0 \pmod{8}.$

Similarly, if $xy+1 \equiv 0 \pmod{3},$ then one of these must be true:

$\begin{cases}\begin{aligned} (x,y) &\equiv (1,2) \pmod{3} \\ (x,y) &\equiv (2,1) \pmod{3} \end{aligned}\end{cases}$

In either of these cases, $x+y \equiv 0 \pmod{3}.$

We've shown that if $xy+1\equiv 0 \pmod{24},$ then $x+y \equiv 0 \pmod{8}$ and $x+y \equiv 0 \pmod{3}.$ It follows that $x+y \equiv 0 \pmod{24}.$

If $xy + 1$ is a multiple of 24, then $xy \equiv -1$ mod 24.

Therefore $x,y$ belong to the multiplicative group modulo 24, $\{ \pm 1, \pm 5, \pm 7, \pm 11 \}$ .

It is easily verified that for any element $z$ in this group, $z^2 \equiv 1$ .

Therefore $xy \equiv -1$ implies $x \equiv -1/y \equiv -y$ and $x + y \equiv 0$ .

Since $xy + 1$ is divisible by $24$ , $xy + 1 = 24n$ for some integer $n$ , and $y = \frac{24n - 1}{x}$ . From this we know that in order for $y$ to be an integer, $x$ cannot be multiple of $2$ or $3$ .

Since $y = \frac{24n - 1}{x}$ , $x + y = x + \frac{24n - 1}{x} = \frac{x^2 - 1 + 24n}{x}$ . Then for each $x \equiv a \pmod{24}$ for $a$ is not a multiple of $2$ or $3$ and $0 < a < 24$ , $x + y \equiv 0 \pmod{24}$ . (For $a = 1$ , $1^2 - 1 = 0 \equiv 0 \pmod{24}$ , for $a = 5$ , $5^2 - 1 = 24 \equiv 0 \pmod{24}$ ; for $a = 7$ , $7^2 - 1 = 48 \equiv 0 \pmod{24}$ ; for $a = 11$ , $11^2 - 1 = 120 \equiv 0 \pmod{24}$ ; for $a = 13$ , $13^2 - 1 = 168 \equiv 0 \pmod{24}$ ; for $a = 17$ , $17^2 - 1 = 288 \equiv 0 \pmod{24}$ ; for $a = 19$ , $19^2 - 1 = 360 \equiv 0 \pmod{24}$ ; for $a = 23$ , $23^2 - 1 = 528 \equiv 0 \pmod{24}$ .)

Therefore, $x + y$ must also be divisible by $24$ .

Let's drop an intuitive solution rather than a mathematical proof. In this question, we only care about the multiplication table for 24. Let's list it out! [0, 24, 48, 72, 96, 120...]

In that table, the only solution for which X and Y can be integers (in xy + 1 = 0 (mod24)) must necessarily have a unit digit of 6.

The reason for this is that for xy + 1 = [0, 24, 48, 72, and 120], no integer solution can be found for xy + 1.

It's simple to show that for xy+1 = 96, the only possible solution is (x,y) = (19,5).

19 + 5 = 24. Woah!

This also follows for 216, and for every other multiple of 24 that has a unit integer of 6. Since the multiples of 24 with a unit integer 6 is the only set of answers that have integer solutions for (x,y), the answer is necessarily Yes, Always.

xy+1=24 xy=23 x=1,23 y=23,1 x+y=24 24 is divisible by 24 and everything stems of off xy=23.

if xy + 1 is congruent to 0 mod 24 then x.y is congruent to -1 mod 24

that means that x is congruent to 1 mod 24 and y is congruent to -1 mod 24

that means that x = 24k+1 and y = 24k-1

which gives us: x + y = 2. 24k which is divisible always by 24

If $xy+1 \equiv 0 \pmod{24}$ , then $xy \equiv -1 \pmod{24} \equiv 23 \pmod{24}$ .

If $a_1 \equiv b_1 \pmod{n}$ and $a_2 \equiv b_2 \pmod{n}$ , then $a_1 a_2 \equiv b_1 b_2 \pmod {n}$ (compatibility with multiplication)

Because 23 is a prime, the only possible combination of 23 is $23 \times 1$ .

i.e. $xy \equiv 23 \times 1 \pmod{24}$ .

Therefore, $x + y \equiv 23 + 1 \pmod {24} \equiv 24 \pmod {24} \equiv 0 \pmod {24}$ .

We have xy=-1 mod 24. This means x and y are odd. Also, this equation means that x and y are not divisible by 3 as xy+1 is. Then we add y^2 on both sides, i.e. y(x+y) = y^2 - 1 = (y+1)(y-1) mod 24. If (y+1)(y-1)=0 mod 24 then x+y=0 mod 24 as y is not. It suffices to show that (y+1)(y-1) is divisible 3 times by 2 and 1 time by 3. It is easily seen that, as y is odd, either y-1 or y+1 is divisible by 4 and the other is divisible by 2 only. This makes it for the "3 times by 2". Finally, one of the element of the triplet (y-1,y,y+1) should be divisible by 3. As y is not, one of y-1 or y+1 is.

If $24|xy+1$ , this means that $xy \equiv 23 \mod 24$ . Because $23$ is a prime, ${x,y}$ have to be ${1,23}$ in any order. The sum of $1$ and $23$ is $24$ , and thus the answer is TRUE

Given that (x+1)(y+1) = (xy+1) + (x+y), if we can show that (x+1)(y+1) is a multiple of 24, then so is x+y.

xy = 2 (mod 3) so either x=1 and y=2 or x=2 and y=1 (all mod 3); no other product will give 2 Hence either x+1 or y+1 = 0 (mod 3)

xy = 3 (mod 4) so either x=1 and y=3 or x=3 and y=1 (all mod 4); no other product will give 3 Hence either x+1=2 and y+1=0 or vice versa. In other words, one of x+1 and y+1 is an even number and the other is a multiple of 4.

The product (x+1)(y+1) therefore includes a multiple of 3, an even number and a multiple of 4. Hence it is divisible by 24. Hence x+y = (x+1)(y+1) - (xy+1) is also divisible by 24.

24|xy+1 Then, xy=24k-1 Or xy=24k+23[I am here just discussing about the form] As product of two integers is off the form 24k+23 then either one of the integers is of the form 24k+23 and the other 24k+1 Which implies x+y =24k Thus x+y is divisible by 24

xy+1=24 xy=23 Factors of 23: 1,23 So x+y=24, thus x+y is always divisible by 24

Consider the following properties

• 1) xy+1 is divisible by 24
• 2) x+y is divisible by 24

Call a pair x,y "valid" if it observes both 1 and 2, and "suitable" if it observes 1. We must prove that every suitable pair is valid.

Notice that, if x,y is suitable, then neither x nor y can have a common divisor with 24 (other than 1), that is, neither of them can be a multiple of 2 or 3. Otherwise, the product of x and y will always be at a multiple of three or two from the next multiple of 24, implying that xy+1 is not equal to 24.

Note that, if x,y is a valid (or suitable) pair, then (x+24),y is also valid (or suitable), since (x+24)y+1=xy+1+24y and (x+24)+y=x+y+24. Analogously, (x-24),y is also a valid (or suitable) pair. Therefore, if every suitable pair x,y with |x|,|y| < 24 is valid, then every suitable pair is valid. Furthermore, if x>-24 is negative, then x+24 is positive, implying that, if every suitable pair x,y with 0<x,y<24 is valid, then every suitable pair is valid. It therefore suffices to prove that every suitable pair x,y with 0<x,y<24 is valid.

Furthermore, let v,w,x be three distinct integers 0<v,w,x<24. We claim that, if v,w is a suitable pair, then v,x is not. Suppose otherwise: if vx were a suitable pair, then vx=vw+24j for some non-zero integer j, implying that either v has a common divisor with 24 (other than 1), or x=w+24n for some integer n, both leading to a contradiction. Therefore, for every positive integer x<24, there is at most one other positive integer y<24 such that x,y is a suitable pair.

The possible values of 0<x<24 for there to be a 0<y<24 such that x,y are a suitable pair are the following: 23,19,17,13,11,7,5,1, all odd numbers between 1 and 24 that are not multiples of 3.

• For x=23, the suitable pair is 23,1, which is valid (23*1+1=24, 23+1=24).
• For x=19, the suitable pair is 19,5, which is valid (19*5+1=96, 19+5=24).
• For x=17, the suitable pair is 17,7, which is valid (17*7+1=120, 17+7=24).
• For x=13, the suitable pair is 13,11, which is valid (13*11+1=144, 13+11=24).

Every other possible value of x has the same suitable pair as those first four.

Therefore, every suitable pair is valid, q.e.d.

If xy = -1 (mod 24), then neither x nor y is divisible by 2 or 3. Then the only possibilities for x are 1, 5, 7, 11, -11, -7, -5, -1. It can be easily verified that the only y-pairings at x=1, 5, 7, or 11 are -1, -5, -7, and -11, respectively. Then x+y = 0 always.

Here's a proof that it AT LEAST sometimes works, but it's not a rigorus proof for the "Yes, always" option.

I observed that if $x=1$ and $y=24n-1$ , then $xy+1=1\cdot(24n-1)+1=24n-1+1=24n$ , and $x+y=1+(24n-1)=24n+1-1=24n$ .

This means that for all values of $24n$ , you can always write it as $xy+1=x+y=24n$ , but i only proved that for $x=1$ and $y=24n-1$ and not the other ways of factoring $24n-1$ . My solution only works for when $24n-1$ is a prime number and for only a single case of when it's a composite number.

An example of $xy+1=24a$ and $x+y=24b$ where x and y is not $1$ and $24n-1$ is $x=7$ and $y=17$ . This one works out too, as $7+17=24$ and $7\cdot17+1=120=24\cdot5$ .

x+y = (x+1)*(y+1) - (xy+1), so it is enough to prove 24 divides (x+1)(y+1)

xy = k*24-1, so

xy mod 3 = 2, so (x,y)=(1,2) mod 3

xy mod 4 = 3, so (x,y)=(1,3) mod 4

so (x+1)(y+1) = (2)mod4 * (0) mod 4 = multiple of 8

and (x+1)(y+1) = (2)mod3 * (0)mod 3 = multiple of 3

q.e.d

Because $x^2 \equiv 1 \pmod{3 ~,~8}$ ( for x, y ( coprime with 24) )
( ∵ for odd number $x=2k+1 ~:~ x^2 = 4k^2 + 4k +1 = 4k(k+1)+1 \equiv 1 \pmod{8}$ , $x= 3k \pm1 ~:~ x^2 \equiv 1 \pmod{3 }$ )

If $xy \equiv -1 \equiv -x^2 \pmod{3 ~,~8}$ , $x+y \equiv 0 \pmod{3 ~,~8}$

∴ x+y is divisible by 24