, 8 tweets, 6 min read
My Authors
Read all threads
@GWOMaths @MathsTim All "congruence" equations are mod 23.

Expanding gives
2x^2 - 5x -3 ≅ 7

or
2x^2 - 5x - 10 ≅ 0

and equivalently
2x^2 - 5x -10 - 23n = 0 for some small n

To be factored requires that for our n, find two integers with difference
5
and product
2.(-10 -23n)
@GWOMaths @MathsTim Suppose n=0:
Require two integers with difference
5
and product
-20.

None found

Suppose n=1:
Require two integers with difference
5
and product
-66.

6 and -11 work.

Back calculating in
2x^2 - 5x - 10 -23n ≅ 0

gives
2x^2 - 5x - 33 ≅ 0

or
(2x - 11) (x + 3) ≅ 0
@GWOMaths @MathsTim For a +ve solution check the first factor
2x - 11 ≅ 0

means that
2x - 11 = 23 i for some i

i=0 gives no integer solution

i=1 gives x = 17 (as 2.17 = 23 + 11)

So x = 17 is a solution.

Verifying:
2 . 17^2 - 5 . 17 - 3
= 2.289 - 85 - 3
= 578 - 88
= 490
= 21. 23 + 7
≅ 7 mod 23
@GWOMaths @MathsTim The second factor yields
x + 3 ≅ 0

equivalently
x + 3 = 23

and
x = 20

Verifying:
2 . 20^2 - 5 . 20 - 3
= 800 - 100 - 3
= 697
= 30.23 + 7
≅ 7 mod 23.

Of our two solutions 17 and 20, the smallest is 17.
@GWOMaths @MathsTim I can't quote the precise relevant theory that these two roots for our equation are unique. (My group theory is very rusty.)

However the residues of 23 with normal addition and multiplication form a Galois field, and thus is in a sense irreducible, leading me to believe it true.
@GWOMaths @MathsTim I have been lazy with signs and descriptions above.

More precisely, integer factorization of
a.x^2 - b.x - c

requires finding two integers s and t, of opposite signs, such that
s + t = -b
and
s . t = - a . c

In our precise case, such that
s + t = -5
and
s . t = -66
Missing some Tweet in this thread? You can try to force a refresh.

Enjoying this thread?

Keep Current with P. Geerkens ✝️

Profile picture

Stay in touch and get notified when new unrolls are available from this author!

Read all threads

This Thread may be Removed Anytime!

Twitter may remove this content at anytime, convert it as a PDF, save and print for later use!

Try unrolling a thread yourself!

how to unroll video

1) Follow Thread Reader App on Twitter so you can easily mention us!

2) Go to a Twitter thread (series of Tweets by the same owner) and mention us with a keyword "unroll" @threadreaderapp unroll

You can practice here first or read more on our help page!

Follow Us on Twitter!

Did Thread Reader help you today?

Support us! We are indie developers!


This site is made by just three indie developers on a laptop doing marketing, support and development! Read more about the story.

Become a Premium Member ($3.00/month or $30.00/year) and get exclusive features!

Become Premium

Too expensive? Make a small donation by buying us coffee ($5) or help with server cost ($10)

Donate via Paypal Become our Patreon

Thank you for your support!