Prove that 11n^2 - 7 is divisible by 4 for all odd integers

In this video I do a divisibilty proof. I prove that 11n^2 - 7 is divisible by 4 for all odd integers. I hope this helps someone.

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[davidhamelehh]

Hey The Math Sorcerer, how has been your week so far?
just reminding you about the video i asked you to make😉
thx for you time, and for all the videos you made, you really help me a ton!

[pandabeargogo]

Reminds me of 1st Uni maths (in UK)!! Not sure if you have done a problem like this on your channel specifically, but, for example, prove that 12^n - 5^n is divisible by 7 for all n (integers) is what I associated this video with.

Of course, one might immediately notice, it sounds like a very obvious result... any 2 different numbers 'to the n' and show it' divisible by...yeah the difference... so 8^n-5^n is divisible by 3 as another example.

For those thinking 'yeah that sounds obvious' Prove it! (by induction). Also, if anyone is confused with Proof by induction procedure. It's 3 steps.

1) Base case (n=1/first value)
2) Hypothesis step (Assume for n=k)
3) Induction Step (Show for n=k+1)

I've seen just now a lot of results just use 2 steps and include n=k assumption as part of the induction step.

A further challenge is proving exactly this...so prove a^n -b^n is divisible by all (a-b) where a, b and n are integers....I recall doing this years ago and I think somewhere you indeed have to remember you can't have non-zero a, b....I say this because if you consider "a^n -b^n is divisible by all (a-b)" you might be tempted so plug in any real number you want...including 0, which you will eventually see in the induction step, you caaant divideee by zzeeero...

Without sounding like I'm losing my grasp on my ability to communite using proper English, I was wondering if you could say for n being any real number. i have just googled 6^(2.6) - 2^(2.6)=99.42275, which isn't divisible by 4^2.6... in fact, using logs, this problem is divisible by 5.86494(...)^2.6, if you want the power value of 2.6 to remain, or, if you want the 'keep there 4 around', which I do, you have 6^2.6 - 2^2.6 | 4^3.31775... which isn't nice and neat at all.

I haven't look at any maths I did over the 3 years at uni in years, so it's exciting to revisit problems and explore the type of thinking that's almost naturally conjured when considering a problem.

I know this 2nd half is incredibly messy, but I wanted to type it out, even for myself.

This also encourages me to look at Proof by induction for factorials again. 2k!, 2k!!, (k+1)! and all that jazz.

... I've just realised after the above this isn't a video for Proof by induction like I immediately assumed it was when I saw the title lmao... Sorry you had to encounter my verbal mess of a comment, but I am fully committed to posting my thoughts about the above

[moirai]

Nice little proof. Instructive on how to think about these things. I thought you were going to reduce the original expression modulo 4 first but I was overthinking it.

[abcforlife]

amazing timing im working through this type of proof right exactly now, thanks math sorcerer

[kqnrqdtqqtttel]

Another way: If we consider the 11n^2 - 7 mod 4, we get 1 - n^2 (mod 4). Notice that the only quadratic residues mod 4 are 0 and 1, but n is not divisible by 2 by assumption, so it must hold that n^2 is congruent to 1 mod 4. We then get the desired result.

[sportmaster]

Is this A Level or university level math ?