Prove that 11^n - 4^n is divisible by 7 for any natural number, n. [Mathematical Induction]

In this video, I showed how to integrate cscx and then use trig identities to obtain other forms of the antiderivative.

[@ODOk-vs8ut]

I shall give a quick method using theory of binomial (a+b)^n= a^n+nC1a^(n-1)b+.............+nCn b^n So we can write 11^n=(7+4)^n Now problem reduces to- (7+4)^n-4^n =>4^n+nC1(4^n-1)[7]+.......+7^n -4^n So we 4^n gets cancelled and thereby every term left containes a factor of 7. HP

[@davidbrisbane7206]

Very good!

[@matheusjahnke8643]

Alternatively: aⁿ-bⁿ=(a-b)(aⁿ⁻¹ + aⁿ⁻²b¹ + aⁿ⁻³b² + .... +a¹bⁿ⁻²+bⁿ⁻¹) You can remember that as a notable product(just like you know (a+b)ⁿ is a binomial).... alternatively. let's take S=aⁿ⁻¹ + aⁿ⁻²b¹ + aⁿ⁻³b² + .... +a¹bⁿ⁻²+bⁿ⁻¹ This is a geometric progression with factor b/a and n terms.... using the geometric sum formula: S=aⁿ⁻¹ (1 - (b/a)ⁿ) / (1 - (b/a)) S=a(aⁿ⁻¹ - bⁿ/a) / (a - b) S=(aⁿ - bⁿ) / (a - b) Alternatively, using the geometric sum trick, take S and multiply by (b/a) (b/a)S = aⁿ⁻²b¹ + aⁿ⁻³b² + .... +a¹bⁿ⁻²+bⁿ⁻¹+bⁿ/a Subtract from S: S - (b/a)S = (aⁿ⁻¹ + aⁿ⁻²b¹ + aⁿ⁻³b² + .... +a¹bⁿ⁻²+bⁿ⁻¹)-(aⁿ⁻²b¹ + aⁿ⁻³b² + .... +a¹bⁿ⁻²+bⁿ⁻¹+bⁿ/a) Note how almost all terms in the RHS are the same, except aⁿ⁻¹ and bⁿ/a S - (b/a)S = aⁿ⁻¹-bⁿ/a (Lore: the induction is hidden on this step... this is a telescoping sum) (1-(b/a)) S = (1/a)(aⁿ-bⁿ) S = (1/a)(aⁿ-bⁿ) / (1-(b/a)) If you arrange it out you should arrive at the same expression back again. The geometric sum fails at a=0... in this case aⁿ-bⁿ=(a-b)(aⁿ⁻¹ + aⁿ⁻²b¹ + aⁿ⁻³b² + .... +a¹bⁿ⁻²+bⁿ⁻¹) Becomes: bⁿ=(-b)(bⁿ⁻¹) And there also a division by 0 induced when a=b... in this case aⁿ-bⁿ=(a-b)(aⁿ⁻¹ + aⁿ⁻²b¹ + aⁿ⁻³b² + .... +a¹bⁿ⁻²+bⁿ⁻¹) Becomes aⁿ-aⁿ=(a-a)(....) 0=0 Which also checks out(Bonus lore: you can use that to prove the derivative of xⁿ is nxⁿ⁻¹ if you are careful on this case)

[@polsiv]

Great work as always!

[@PrimeNewtons]

Thanks

[@user-gm8xo2ug3o]

This has really helped! A big thank from South Africa!

[@EzraSchroeder]

this one is straight out of an advanced calculus book haha! it is a problem in chapter one of Kenneth A. Ross: Analysis the Theory of Calculus -- love it!

[@jonathancollett4174]

Awesome and clear video!! I did not even think of breaking down that number into two different numbers.

[@customperforma6800]

wow, I have exams tomorrow. You just saved me. BTW we are from Sydney!!

[@PrimeNewtons]

Hello from LA

[@bhaveerathod2373]

I tried on my own but got stumped after 7:15, thank you for your wonderful trick!

[@EzraSchroeder]

so good--so satisfying!

[@Why553-k5b_1]

for even numbers, it's divisible by 7 (because 11²-4²=(7)(15) and 11⁴-4⁴=(121-16)(121+16)=(105)(137)=(7)(15)(137)) and for numbers that are divisible by 3, it's also divisible, but i don't know how with primes (sorry for bad english)

[@davidhardy9419]

A^n - B^n is always divisible by A-B. Even n is obvious, odd n more difficult.

[@lisakhanyamashiya1428]

Thank you sir

[@RYedukrishnan-cn5ft]

Hello sir ❤, thanks for this video I have a doubt, 'm' will never be a -ve number and zero. Then why are you saying 'm' is an integer 😂. I will say 'm' is a natural number. If you say that both are correct then I will say 'm' is a Complex numbers( set of all kinds of numbers 😂😂😂😂😂). If you say a natural number then problem solved. Other wise it is a debatable topic 😂😂. anyway thank you sir for this video. My poor English hop you understand 😁

[@oarabilemoitshela444]

so helpful, thank you!!!!

[@StuartSimon]

It is even true (trivially) for n=0.

[@marcgriselhubert3915]

11 is congruent to 4 mod 7, so 11^n is congruent to 4^n mod 7 and 11^n - 4^n is congruent to 4^n - 4^4 = 0 mod 7, and that's finished.

[@moi769]

I used the remarkable identity (a^n -b^n ) to prove that ( 11^n - 4^n) is a multiple of 7 so we can wright it as 7*k wich k is a natural number

[@dan-florinchereches4892]

I see. It makes sense to to relate a^n-b^n=(a-b)(a^(n-1)+a^(n-2)b+...+b^(n-1)) I wanted to use binomial expansion by writing 11^n-4^n=(7+4)^n-4^n = choose(n,k)*7^(n-k)*4^k -4^k . So 4^n cancel out and we are left with all the terms for k from 0 to n-1 which all contain a multiple of 7. Qed My thought went to a problem I knew before: Show that fraction (2^(n+3) + 3^n*5^(n+1))/(2^(n+2)×3^(n+1)+5^(2n)) Is always reductible by 13

[@solethumatrose]

Wow

[@vemarj2802]

Also: 2^(2(2n+1)) + 1 is divisible by 5 for all n >= 0 2^(4(2n+1)) + 1 is divisible by 17 for all n >= 0 😊

[@ManojkantSamal]

11-4=7, This is the core reason...

[@elmer6123]

This problem is for natural numbers so a, b, c, and n are natural numbers. If a%c=0 then a^n is divisible by c. If a%c-b%c=0, then (a^n-b^n)%c=0 and a^n-b^n is divisible by c. Example: 11%7=4 and 4%7=4. 11^13=34522712143931 and 34522712143931%7=4; 4^13=67108864 and 67108864%7=4 (11^13-4^13)=34522712143931-67108864=34522645035067 and 34522645035067%7=0, so (11^13-4^13) is divisible by 7.

[@robertveith6383]

Stop using a percent symbol for the intended operation.