# Divisors and polynomials

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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)

"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

(Original post by

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)

**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)

Also 91 has two (prime) factors 7, 13.

Can you take it from there?

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(Original post by

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?

**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?

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#4

(Original post by

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

**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

What equation must exist if x^2-1 is divisible by 91?

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(Original post by

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?

**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?

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#6

(Original post by

I don't know if this is what you mean but there exists a constant say m such that 91m=x^2-1

**MissMathsxo**)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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(Original post by

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 ...

**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 ...

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#8

(Original post by

I realise that I'm probably overthinking this and missing something obvious but i cant see where this is going, sorry🙃

**MissMathsxo**)I realise that I'm probably overthinking this and missing something obvious but i cant see where this is going, sorry🙃

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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(Original post by

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.

**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.

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#10

(Original post by

Am I supposed to be replacing (x+1)(x-1) with 7 and 13?

**MissMathsxo**)Am I supposed to be replacing (x+1)(x-1) with 7 and 13?

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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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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#12

(Original post by

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

**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

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(Original post by

How did you come up with those values? I'm off bed shortly so ...

**mqb2766**)How did you come up with those values? I'm off bed shortly so ...

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#14

(Original post by

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?

**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?

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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