# Divisors and polynomials

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#1
Could someone please help me understand why the following is true and how you work it out.
"There are 4 numbers between 0 and 91 such that 91 divides (x^2)-1".
Likewise can anyone give an example of 2 irreducible polynomials who's sum is reducible.
Thanks (for reference this is for algebra revision)
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2 years ago
#2
(Original post by MissMathsxo)
Could someone please help me understand why the following is true and how you work it out.
"There are 4 numbers between 0 and 91 such that 91 divides (x^2)-1".
Likewise can anyone give an example of 2 irreducible polynomials who's sum is reducible.
Thanks (for reference this is for algebra revision)
For the first one, notice x^2-1 is the difference of two squares (x+1)(x-1)
Also 91 has two (prime) factors 7, 13.
Can you take it from there?
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#3
(Original post by mqb2766)
For the first one, notice x^2-1 is the difference of two squares (x+1)(x-1)
Also 91 has two (prime) factors 7, 13.
Can you take it from there?
Sorry I'm still not sure where to go from here. I just can't work out what part of my course this links to
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2 years ago
#4
(Original post by MissMathsxo)
Sorry I'm still not sure where to go from here. I just can't work out what part of my course this links to
It may surprise you that I've got no idea how it relates to your course, probably more so than you.

What equation must exist if x^2-1 is divisible by 91?
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#5
(Original post by mqb2766)
It may surprise you that I've got no idea how it relates to your course, probably more so than you.

What equation must exist if x^2-1 is divisible by 91?
I don't know if this is what you mean but there exists a constant say m such that 91m=x^2-1
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2 years ago
#6
(Original post by MissMathsxo)
I don't know if this is what you mean but there exists a constant say m such that 91m=x^2-1
Yes. This is "hard" because its a single quadratic in two integer variables. The obvious thing to try is using factors, so sub in the two relationships given earlier ...
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#7
(Original post by mqb2766)
Yes. This is "hard" because its a single quadratic in two integer variables. The obvious thing to try is using factors, so sub in the two relationships given earlier ...
I realise that I'm probably overthinking this and missing something obvious but i cant see where this is going, sorry🙃
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2 years ago
#8
(Original post by MissMathsxo)
I realise that I'm probably overthinking this and missing something obvious but i cant see where this is going, sorry🙃
Pls substitute the two given relationships in post 2 into your equation.

How to spot where its going? It should be easy to see x^2-1 is the difference of two squares, it's shouting at you to use this fact. Therefore the answer has something to do with what are the two factors that give (x+1)(x-1). 91 has two prime factors 7 & 13, again the two factors thing pops up again. At least write down the next line.
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#9
(Original post by mqb2766)
Pls substitute the two given relationships in post 2 into your equation.

How to spot where its going? It should be easy to see x^2-1 is the difference of two squares, it's shouting at you to use this fact. Therefore the answer has something to do with what are the two factors that give (x+1)(x-1). 91 has two prime factors 7 & 13, again the two factors thing pops up again. At least write down the next line.
Am I supposed to be replacing (x+1)(x-1) with 7 and 13?
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2 years ago
#10
(Original post by MissMathsxo)
Am I supposed to be replacing (x+1)(x-1) with 7 and 13?
From post 5
91m = x^2-1
so
7*13*m = (x+1)*(x-1)

What do you notice about the two factors on the right hand side? How can you choose "m" to give you that relationship on the left hand side?
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#11
I could only find 8 and 45 as values of m that work. This only gives me two x values opposed to the 4 I need
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2 years ago
#12
(Original post by MissMathsxo)
I could only find 8 and 45 as values of m that work. This only gives me two x values opposed to the 4 I need
How did you come up with those values? I'm off bed shortly so ...
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#13
(Original post by mqb2766)
How did you come up with those values? I'm off bed shortly so ...
I just spotted them from the difference between the two numbers for example m of 8 would give x of 27 and 28=7*4 and 26=13*2. 2*4=8. Are these values incorrect then?
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2 years ago
#14
(Original post by MissMathsxo)
I just spotted them from the difference between the two numbers for example m of 8 would give x of 27 and 28=7*4 and 26=13*2. 2*4=8. Are these values incorrect then?
They're right. If the two factors on the right hand side are (x+1) and (x-1), then they differ by 2. Therefore,
7*k1 and 13*k2
must differ by 2 when m = k1*k2. So you're just running through the multiples remainder 2 or modulo 7 or 13 arithmetic.

Your first one is (as you say)
28, 26 where k1=4, k2=2, m = 8, x=27.
Your other one is (as you say)
63, 65 where k1=9, k2=5, m=45, x=64
Note they sum to 91.
The next one would be x=118, but that is >91. Modulo 91, those are the only two solutions.

There is an easy or/ trivial "first" one, and its mirror (sum to 91), which gives the four values. Have a think about them, I'm off bed :-)

Edit: For the irreducible polynomials, are they over the integers, reals, ...?
Last edited by mqb2766; 2 years ago
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