Basic number theory problem

1. May 3, 2012

Gravitational

Let x and y be integers. Prove that 2x + 3y is divisible
by 17 iﬀ 9x + 5y is divisible by 17.
Solution. 17 | (2x + 3y) ⇒ 17 | [13(2x + 3y)], or 17 | (26x + 39y) ⇒
17 | (9x + 5y), and conversely, 17 | (9x + 5y) ⇒ 17 | [4(9x + 5y)], or
17 | (36x + 20y) ⇒ 17 | (2x + 3y)

Could someone please help me understand this solution. I do not understand it at all. What basis do they have for doing such operations? The solution just doesn't make sense

2. May 3, 2012

micromass

Staff Emeritus
Which part don't you understand??

The only two rules they used were

$$n\vert m~\Rightarrow~n\vert mk$$
and
$$n\vert m,~n\vert k~\Rightarrow~n\vert (m+k)$$

3. May 3, 2012

Gravitational

Why do they multiply 2x+3y by 13? and why do they multiply 9x+5y by 4? why not some other numbers?

4. May 3, 2012

Hurkyl

Staff Emeritus
So they can get the desired answer. If you use other numbers, you'll get other equivalent statements.

How did they know that would give the desired answer? Trial and error would work. Or, you could try and write down an equation that says "If I multiply by n, then the answer I get is the one I want".

You're asking the wrong question, it seems. You don't seem to have meant "I don't understand this solution!" -- you seem to have meant "How could I have come up with this solution myself?"

5. May 3, 2012

micromass

Staff Emeritus
Do you know modulo arithmetic?

6. May 3, 2012

Gravitational

no, but i dont think modulo arithmetic is necessary in this problem

7. May 3, 2012

Gravitational

I understand why they multiplied by 13, but i dont see the significance in multiplying by 4

8. May 3, 2012

micromass

Staff Emeritus
It's not necessary in understanding the solution. But it's necessary in understanding why they did what they did.

9. May 3, 2012

ramsey2879

If you understand the first part, then you should understand the logic of the second part. Both parts are required to show the "if and only if" condition. Basically, they used multiplication by 4 since 4*9 = 17*2 plus 2 and that is the way to reduce the 9x to 2x mod 17. Because of the iff part, taking care of the x variable also takes care of the y variable.

Last edited: May 3, 2012
10. May 6, 2012

Gravitational

2x + 3y is divisible by 17, there is an integer k such that (2x + 3y)/17 = k <=> 2x + 3y = 17k. Multiply both sides by 13
13(2x + 3y) = 13 * 17k
<=>
26x + 39y = 13 * 17k
<=>
9x + 5y + (17x + 34y) = 13 * 17k
<=> (moving over the thing in the parantheses to the right-hand side and factoring out 17
9x + 5y = 13 * 17k - (17x + 34y) = 13 * 17k - 17(x + 2y) = 17(13k - (x + 2y))

Thus 9x + 5y is divisible by 17.

I actually think this proof is better

11. May 6, 2012

Gravitational

solution*

12. May 6, 2012

ramsey2879

More powerful if you use the Mod operations. In Mod 17, multiples of 17 == 0.

2x + 3y == 0 Mod 17
26x + 39y == 13* 0 Mod 17
9x + x*0 + 5y + y*0 == 13*0 Mod 17
9x + 5y == 0 Mod 17

I other words 9x+ 5y is divisible by 17 if 2x + 3y is divisible by 17