MAT question help

I tried the 2022 MAT paper and got all the multiple choice questions correct except this one:

Spoilers below if those reading this want to attempt it :smile:

I spent a while looking at the given factorisation and trying to work out how it connects to when n^4+4 is prime but in the end I gave up (I only knew it wasn't (a) because of n=1). The solution is to make a substitution n=m/sqrt(2) which makes sense but I'm not sure I would have spotted that in a reasonable time.

If I wasn't given the factorisation and just asked about n^4+4 then I would have attempted to factorise it and I don't think it's too hard to realise that it's (n^2 + 2n + 2)(n^2 - 2n + 2). Then I'm confident that I would have got the answer from there. I should have tried factorising n^4+4 but I was too focussed on trying to use the given result.

So I'm wondering what the point of giving the n^4+1 factorisation is because it doesn't seem necessary and was a big distraction, for me at least? Is this an "obvious" technique that Oxbridge students would be expected to see?

(edited 9 months ago)

Reply 1

mqb2766

21

Original post by 0-)

I tried the 2022 MAT paper and got all the multiple choice questions correct except this one:

Spoilers below if those reading this want to attempt it :smile:

I spent a while looking at the given factorisation and trying to work out how it connects to when n^4+4 is prime but in the end I gave up (I only knew it wasn't (a) because of n=1). The solution is to make a substitution n=m/sqrt(2) which makes sense but I'm not sure I would have spotted that in a reasonable time.

If I wasn't given the factorisation and just asked about n^4+4 then I would have attempted to factorise it and I don't think it's too hard to realise that it's (n^2 + 2n + 2)(n^2 - 2n + 2). Then I'm confident that I would have got the answer from there. I should have tried factorising n^4+4 but I was too focussed on trying to use the given result.

So I'm wondering what the point of giving the n^4+1 factorisation is because it doesn't seem necessary and was a big distraction, for me at least? Is this an "obvious" technique that Oxbridge students would be expected to see?

I would have taken it as a hint about how to factorise n^4+4, and a hint that you shoud do it. So factorise it as you say as (n^2 + 2n + 2)(n^2 - 2n + 2). The deduce is a bit of a hint for the substitution and it shouldnt be too hard to spot that
n^4+4 = 4(n^4/4+1) = 4(m^4+1)
where m=n/sqrt(2) as you say, but Id prefer it without the sqrt(2)s in the factorisation.

(edited 9 months ago)

Reply 2

0-)

OP

11

Original post by mqb2766

I would have taken it as a hint about how to factorise n^4+4, and a hint that you shoud do it. So factorise it as you say as (n^2 + 2n + 2)(n^2 - 2n + 2). The deduce is a bit of a hint for the substitution and it shouldnt be too hard to spot that
n^4+4 = 4(n^4/4+1) = 4(m^4+1)
where m=n/sqrt(2) as you say, but Id prefer it without the sqrt(2)s in the factorisation.

I didn't consider that this was a hint for factorising n^4+4 but maybe that's just me. I thought there didn't need to be any more algebra and I could solve the problem just using the n^4+1 factorisation and considering the "+3" of similar but that didn't lead to anything.

To me the question is simpler without the hint than with it which makes me feel like it's not a very good question. But again maybe I'm in the minority.

Reply 3

DFranklin

18

Original post by 0-)

I didn't consider that this was a hint for factorising n^4+4 but maybe that's just me. I thought there didn't need to be any more algebra and I could solve the problem just using the n^4+1 factorisation and considering the "+3" of similar but that didn't lead to anything.

To me the question is simpler without the hint than with it which makes me feel like it's not a very good question. But again maybe I'm in the minority.

FWIW, I agree with mqb that it's a hint that x^4+4 factorizes and what the factorization "looks like". Only difference is that I'd look at the given factorization and see "how it works" and then adjust it to get a factorization of x^4+4.

Perhaps worth noting that finding

\int \dfrac{1}{x^4+1}\,dx

is a classic "difficult integral" which relies on the given factorization so *some* candidates would probably be aware of it.

Reply 4

mqb2766

21

Original post by 0-)

I didn't consider that this was a hint for factorising n^4+4 but maybe that's just me. I thought there didn't need to be any more algebra and I could solve the problem just using the n^4+1 factorisation and considering the "+3" of similar but that didn't lead to anything.

To me the question is simpler without the hint than with it which makes me feel like it's not a very good question. But again maybe I'm in the minority.

If youre showing that the polynomial isnt prime almost always, youre usually looking for an integer factorisation so +3 isnt going to get you anywhere and the given factorisation for n^2+1 isnt that useful for considering primes as it contains a sqrt(2) so the product for n^2+1 will be two surds which isnt what youre looking for. Tbh the question is similar to something like determine when
a^2 - b^2
is prime. So using dots (a+b)(a-b) its usually composite apart from when a-b=1, so consecutive numbers, and a+b is prime.

Id have gone down the route of the "new" factorisation and taken the given one as a hint. It seems simpler than arguing about a scaling by a surd scale factor, though thats just me. If you interpret "deduce" as a surd scaling factor you still would probably have to stick it into the given expression and verify when it gives a product of integers (not hard, but its still a line or two of algebra).

Even without the hint you could notice that n^4+4 is ~quadratic so you "always" ~complete the square so
(n^2+2)^2 - (2n)^2
which, using dots, gives the factorisation with integer coeffs, so n=1 is the only case where one of the factors is 1 and the other is 5.

(edited 8 months ago)

Reply 5

DFranklin

18

Original post by mqb2766

If youre showing that the polynomial isnt prime almost always, youre usually looking for an integer factorisation so +3 isnt going to get you anywhere and the given factorisation for n^2+1 isnt that useful for considering primes as it contains a sqrt(2) so the product for n^2+1 will be two surds which isnt what youre looking for. Tbh the question is similar to somehting like determine how many times
a^2 - b^2
is prime. So using dots (a+b)(a-b) its usually composite apart from when a-b=1, so consecutive numbers, and a+b is prime.

Id have gone down the route of the "new" factorisation and taken the given one as a hint. It seems simpler than arguing about a scaling by a surd scale factor, though thats just me. If you interpret "deduce" as a surd scaling factor you still would probably have to stick it into the given expression and verify when it gives a product of integers (not hard, but its still a line or two of algebra).

Even without the hint you could notice that n^4+4 is ~quadratic so you "always" complete the square so
(n^2+2)^2 - (2n)^2
which, using dots, gives the factorisation with integer coeffs, so n=1 is the only case where one of the factors is 1 and the other is 5.

I think the idea is "use the scaling factor to get a factorization of n^4+4"; once you've done that you don't really need to worry about surd multipliers etc. At the same time it's not terribly obvious it will work out that way and I'd agree it's fairly counter intuitive to scale by a surd when you're looking for integer factors. (Although arguably you'd realise you *have* to scale by sqrt(2) if you want the given factorization to yield integers).

Reply 6

mqb2766

21

Original post by DFranklin

I think the idea is "use the scaling factor to get a factorization of n^4+4"; once you've done that you don't really need to worry about surd multipliers etc. At the same time it's not terribly obvious it will work out that way and I'd agree it's fairly counter intuitive to scale by a surd when you're looking for integer factors. (Although arguably you'd realise you *have* to scale by sqrt(2) if you want the given factorization to yield integers).

Agreed, thats what I was trying to say. The direct factorisation seems more obvious than working through a surd scaling factor for a surd expression, though obv they end up the same.

Reply 7

PAT_22

12

Original post by 0-)

I tried the 2022 MAT paper and got all the multiple choice questions correct except this one:

Spoilers below if those reading this want to attempt it :smile:

I spent a while looking at the given factorisation and trying to work out how it connects to when n^4+4 is prime but in the end I gave up (I only knew it wasn't (a) because of n=1). The solution is to make a substitution n=m/sqrt(2) which makes sense but I'm not sure I would have spotted that in a reasonable time.

If I wasn't given the factorisation and just asked about n^4+4 then I would have attempted to factorise it and I don't think it's too hard to realise that it's (n^2 + 2n + 2)(n^2 - 2n + 2). Then I'm confident that I would have got the answer from there. I should have tried factorising n^4+4 but I was too focussed on trying to use the given result.

So I'm wondering what the point of giving the n^4+1 factorisation is because it doesn't seem necessary and was a big distraction, for me at least? Is this an "obvious" technique that Oxbridge students would be expected to see?

The ‘scaling’ idea is a sort of manifestation of ‘homogenous polynomials’ I think. Scale a homogeneous polynomial and the polynomial will scale along with it. This doesn’t necessarily work for all polynomials, but the fact that different terms scale differently depending on their degree can be super helpful I guess. Just gotta learn these tricks.
This factorization is known as the Sophie Germain identity.

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