Fido Johnson

Number theorist.

[Fido Johnson @JohnsonFido]

@GWOMaths Today we enjoy the famous MARKOV EQUATION

x² + y² + z² = 3xyz

This simple equation results in some utterly fabulous number theory!

Fancy some fun? Come take a peek.

[Thread…]

[Fido Johnson @JohnsonFido]

@twarrestrial @mathhub_vn I tried that line for a bit and got nowhere. So I did this:

[Fido Johnson @JohnsonFido]

@chris_juravich @jamestanton My pleasure.

The solution obviously generalises to give m consecutive integers each with n distinct prime divisors. I hope I haven’t ruined the game 😂

[Fido Johnson @JohnsonFido]

@almatematico86 @jamestanton The question only requires that the nine primes be distinct.

So you may have two even numbers in your sequence as long as there are sufficiently many other distinct primes.

[Fido Johnson @JohnsonFido]

@chris_juravich @microfusio91945 They didn’t have much choice. 1/7 has a unique non trivial EF of length 2 😜

[Fido Johnson @JohnsonFido]

@chris_juravich @microfusio91945 Most obviously, restrict the length of the Egyptian fraction to 3 from 4. This will drastically reduce the number of solutions, although there may still be dozens of them…

[Fido Johnson @JohnsonFido]

@t0tientqu0tient Particularly in the last 20 years, too much proof writing is from the perspective of ‘let me show you how clever I am’ whereas it should be ‘let me explain this to you’.

But also, a pair of well worn shoes are often more comfortable than new ones.

[Fido Johnson @JohnsonFido]

@Math_files Let me guess, you divided by zero or forgot that √x²=±x.

[Fido Johnson @JohnsonFido]

@ihtesham2005 So it’s fair enough to praise AoPS as great preparation for maths contests in particular, and also for some general problem solving utility.

But a comment like ‘almost every existing math textbook was teaching the wrong thing’ is frankly ridiculous.

[Fido Johnson @JohnsonFido]

@ihtesham2005 ‘Competition maths’ is a particular branch of mathematics whereby you try to solve bite sized problems under time pressure. Some of the winning techniques are very limited, so much so that they rarely if ever show up anywhere other than maths competitions.

[Fido Johnson @JohnsonFido]

@nazirafzal I don’t think this is a wise post.

[Fido Johnson @JohnsonFido]

@chris_juravich @microfusio91945 I don’t think so, no.

[Fido Johnson @JohnsonFido]

@chris_juravich @microfusio91945 To see this, note that allowing ± ℤ is equivalent to solving in +ℤ one of:
1/A + 1/B + 1/C = 1/7 + 1/D
1/A + 1/B = 1/7 + 1/C + 1/D
1/A = 1/7 + 1/B + 1/C + 1/D

But RHS > 1/7, and hence the same bounding argument as before limits the number of choices for the LHS variables.

[Fido Johnson @JohnsonFido]

@chris_juravich @microfusio91945 With negative integers allowed, it’s immediate that there are infinitely many trivial solutions.

For example, (8, 56, n, -n) or (7, 2n, -3n, -6n)

But there are still only finitely many non-trivial solutions, i.e. with
A,B,C,D ≠ 7
|A|, |B|, |C|, |D| distinct

[Fido Johnson @JohnsonFido]

@amancalledprak @RadishHarmers As a retired number theorist I hate to have to tell you but ‘young sheldon’ is a TV character written principally to entertain. He does not typically feature on reading lists.

[Fido Johnson @JohnsonFido]

@podsavetheuk He really is a cretin.

[Fido Johnson @JohnsonFido]

@Urban__J @MrCDraws @Scott__Cheggs If she had said ‘Good morning sir, we’re with a vulnerable person, would you mind using the other side of the road?” he may well have said yes.

Her appalling behaviour is symptomatic of a broader issue with authoritarian policing.

Leadership change is required.

[Fido Johnson @JohnsonFido]

@ZahlenRMD Since P(x) has real coefficients, its non-real roots will appear in complex conjugate pairs.

Hence the roots of P(x) are 1 ± i and 2 ± i.

The constant term in P(x) is the product of its roots.

Hence d = (1+i)(1-i)(2+i)(2-i) = 2 • 5 = 10 😜

[Fido Johnson @JohnsonFido]

@chris_juravich @microfusio91945 So I set myself the pencil and paper task of finding a solution with distinct denominators and containing 1/8.

So far, by playing with 2s, 3s and 7s, I have (12,36,42,126).

Still working on a solution with 1/8…

[Fido Johnson @JohnsonFido]

@chris_juravich @microfusio91945 Footnote: typically for Egyptian fractions we require the denominators to be positive integers. Some texts will also require that denominators be distinct.