Prove 3^n + 7^n -2 is divisible by 4

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  • Опубліковано 27 січ 2025

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  • @Jorje-is-on-fire
    @Jorje-is-on-fire Місяць тому +18

    First. Can i have a pin? I'm your biggest fan and you helped me a lot with my homework

  • @abdoonyoutube7997
    @abdoonyoutube7997 Місяць тому +12

    Man your pacing for teaching stuff is unbelieveable. It would be a literal dream if you would teach olympiad stuff.

  • @timer570
    @timer570 Місяць тому +2

    Hello sir. I enjoy mathematics and I enjoy watching your videos. You are very good at teaching. I also love Jesus and I like that you spread the word. God bless you sir. ❤️

  • @itsphoenixingtime
    @itsphoenixingtime Місяць тому +7

    I tried todo some manipulation.
    3^n + 7^n - 2
    Let us turn the 3^n to 2(3^n) - 3^n
    = 2(3^n) - 3^n + 7^n - 2
    And regroup
    = 7^n - 3^n + 2(3^n) - 2
    Now I want to use an identity where a^n - b^n = (a-b) (a^n-1 + ... + b^n-1)
    Another way to see it is that a-b divides a^n - b^n by using the factor theorem and letting a^n - b^n be a polynomial in a and letting a = b.
    So:
    = (7 - 3)(7^n-1 + .... + 3^n-1) + 2(3^n - 1)
    = (7 - 3)(7^n-1 + .... + 3^n-1) + 2( (3-1) (3^n-1 + ... + 1))
    = 4(7^n-1 + .... + 3^n-1) + 2(2)(3^n-1 + ... + 1)
    = 4(7^n-1 + .... + 3^n-1) + 4(3^n-1 + ... + 1)
    = 4p + 4q
    = 4(p+q) = divisible by 4.
    Hence, it is divisible for n greater than or equal to 0.
    If it's concerning, then we can also just say the identity holds for n >= 1, and then n = 0 is "obvious" because the expression is 0, and 0 divides any number. So it applies for n >= 0

    • @StudyOnly-nn1xb
      @StudyOnly-nn1xb Місяць тому

      But what 0.5 or values between 0 and 1

    • @itsphoenixingtime
      @itsphoenixingtime Місяць тому

      @@StudyOnly-nn1xb I assume they mean integers. If you raise an integer to an irrational/non-integer power, you will most certainly not get an integer and hence there is no meaninful way to talk about "divisible by 4."

    • @StudyOnly-nn1xb
      @StudyOnly-nn1xb Місяць тому

      @@itsphoenixingtime oh I got confused as n€integer or natural wasnt mentioned in question

    • @itsphoenixingtime
      @itsphoenixingtime Місяць тому

      @@StudyOnly-nn1xb I think it is usually assumed that the inputs of n are integers, because divisibility problems only work with integers. For example you could sya that 4^2 and 4^3 are divisible by 4, but you wouldn't say the same for 4^pi ot 4^0.5.
      Either ways, the question SHOULD specific that n is the set of positive integers inclusive of 0, Z+ U {0}.
      Hope this clarifies.

    • @SacMot
      @SacMot Місяць тому

      @@itsphoenixingtimeYou're right, but the problem SHOULD NOT clarify that, it is extremely obvious, just like saying you should put the "2" in the square root. I mean, if you want to do it, no problem, but it's not necessary.

  • @alipourzand6499
    @alipourzand6499 Місяць тому +4

    Great video! The advantage of modular arithmetic approavh is that we can generate infinitely many relations like this:
    4 divides (3+4u)^n + (7+4v)^n - 2
    u, v being integers.

  • @Sphinxinator
    @Sphinxinator 25 днів тому

    This was a really good way of explaining! I was about to use logarithms (I just recently found out about it) but didn't know what to do, so I watched your explanation - and it was very simple!

  • @ziaulhaqqadri631
    @ziaulhaqqadri631 Місяць тому

    I had some trouble with proof by M.I as I'm new to it. I was behind in my class. Luckily I found your videos a while back and now I'm much more confident with proof. You've taught me more than my teachers so I can't thank you enough. Keep making these brilliant videos ❤

  • @krwada
    @krwada Місяць тому +2

    I love the modular proof. Great video.
    By the way, what is the music to your intro? It sounds Jamaican. Very nice, I always enjoy your videos. Math can be fun!

  • @blessingtshuma3760
    @blessingtshuma3760 Місяць тому +1

    Awesome video, would love to see more videos with different modular arithmetic properties, especially for Olympiad mathematics.
    Love from Eswatini 💙

  • @robertlunderwood
    @robertlunderwood Місяць тому +3

    I was trying to avoid using induction here because it was too easy. I like the modular arithmetic here, although that alternative property probably needs an induction proof itself.
    The approach I am trying involves converting 3 to 4-1 and 7 to 8-1 and fiddling with Pascal's triangle.
    Edit: I just figured it out. Every term except the last one for (4-1)^n and (8-1)^n is a multiple of 4. The last term for even n is 1. Taking away 2 leaves only the multiples of 4. The last term for odd n is -1. Taking away 2 leaves a -4 along with the multiples of 4.

    • @PrimeNewtons
      @PrimeNewtons  Місяць тому

      That works too. I've seen that in the comments

  • @MadaraUchihaSecondRikudo
    @MadaraUchihaSecondRikudo Місяць тому +6

    The 7 and 3 being equally congruent mod 4 is easier to visualize when you consider that both 3 and 7 are congruent to -1 (mod 4), which you then take powers of and alternating between 1 and -1, and -1 (mod 4) is equivalent to 3 (mod 4).

    • @sobolzeev
      @sobolzeev Місяць тому +3

      And then you should only observe that +2=-2 mod(4) to conclude the assertion.

    • @bjornfeuerbacher5514
      @bjornfeuerbacher5514 Місяць тому +1

      "when you consider that both 3 and 7 are congruent to -1 (mod 4)"
      I pointed that out already in the comment section to the last video, where the question which is answered here was posted. ;)

    • @sobolzeev
      @sobolzeev Місяць тому +2

      @@bjornfeuerbacher5514 You may also observe that 3² = 7² = 1 mod(8). Hence easily conclude that the expression is in fact divisible by 8, not just 4.

    • @bjornfeuerbacher5514
      @bjornfeuerbacher5514 Місяць тому

      @@sobolzeev Thanks for pointing that out. :)

  • @wannabeactuary01
    @wannabeactuary01 Місяць тому

    Fabulous! Good explanations of both and great revision if I ever needed to use again

  • @spy_44_
    @spy_44_ Місяць тому +18

    Respect from India because in today's time nobody appreciates Maths and Science. It's like nightmare to UA-camrs or reel masters😅!🙏

    • @justnowi8967
      @justnowi8967 Місяць тому +5

      Yes, but don't say nobody, there are few ones.

    • @spy_44_
      @spy_44_ Місяць тому +1

      @justnowi8967 bro/sis/uncle who so ever you are. It wasn't for people like you!🙏

    • @justnowi8967
      @justnowi8967 Місяць тому +2

      @@spy_44_ exactly, like me - who are not blind towards the great Indian people who appreciate, teach and contribute to mathematics, even today, the ones who do not use disrespectful words going to extent saying "nobody appreciates maths & science in india", the ones who try to learn from all people, disregarding nation, gender or any other form of discrimination.

  • @preciousmathematicsfun4331
    @preciousmathematicsfun4331 Місяць тому +1

    I loved the modulus method

  • @張謙-n3l
    @張謙-n3l Місяць тому +1

    MI is a great way to prove this kind of problem. A limitation is that it only works in problems with discrete domains, it never works for problems with continuous domains

    • @abdoonyoutube7997
      @abdoonyoutube7997 Місяць тому

      There is induction for all reals but its a bit more difficult

  • @Mmm---mmm
    @Mmm---mmm Місяць тому +1

    Thank you! Couldn't watch the video earlier i was very occupied with some school stuff but ty

  • @venkateshsundaresan94
    @venkateshsundaresan94 Місяць тому +4

    Another method is to use binomial expansion. Have 3^n as (4 - 1)^n and 7^k as (8 -1) ^k.

  • @dan-florinchereches4892
    @dan-florinchereches4892 Місяць тому +5

    I would go by binomial expansion
    3^n+7^n-2=(4-1)^n+(8-1)^n-2 can't write sigma notation in comments section sadly
    But each term except last one from both binomials is of the form Choose(n,k)*4^(n-k)*(-1)^k or choose(n,k)*8^(n-k)*(-1)^k so everything divides by 4.
    Now what is left for discussion are just the last terms when n=k so we are left to check if (-1)^n+(-1)^n-2 is divisible by 4
    If n=even the term evaluates to 0 divisible by 4
    If N=odd the last term evaluates to -4 which is also divisible by 4 qed

    • @carpvirunga1382
      @carpvirunga1382 Місяць тому +1

      I used binomial expansion as well 3^n+7^n-2=3^n+(4+3)^n-2= 3^n+ n £i=0 (n\k)4^k*3^(n-k)-2= 2(3^n-1)+n£i=1(n/k)4^k*3^(n-k)
      One can write 3^n-1= (3-1) n£i=1*3^(n-i)
      So we can write 4(n£i=1(3^(n-i)+(n/k)4^(k-1)*3(n-k))

  • @alexandermorozov2248
    @alexandermorozov2248 Місяць тому +1

    The second method is interesting, I didn't know about it before. It would be nice to prove more strictly that the remainder of the division of 3^n and 7^n alternates between 1 and 3.

    • @lilellia
      @lilellia Місяць тому

      Since 3 and 7 are equivalent mod 4, their powers will be as well, so we only need to look at one of them. But 3 and 7 are also equivalent to -1 (mod 4) so we just need to look at powers of -1 (mod 4), which clearly alternate between -1 and +1 (mod 4), or equivalently, 3 and 1.

  • @maths01n
    @maths01n Місяць тому

    Watching keenly ❤

  • @Ghosty2k77
    @Ghosty2k77 Місяць тому +1

    Wow, It was really interesting!

  • @SpencerLeDoux-q1q
    @SpencerLeDoux-q1q Місяць тому

    So cool and easy to understand!!

  • @Prypak
    @Prypak Місяць тому

    By using modulo maths, (pretend that the equal sign is actually the congruent sign)
    3=-1[4]
    So
    3^n is congruent to either -1 or 1 depending on the parity of n
    7=-1[4] so 7^n is also congruent to either -1 or 1 depending on the parity of n
    And them together :
    3^n+7^n=-2[4] (either -2 or 2)
    Take away 2 from both sides and for the -2 case, notice that -4=0[4]
    There you go

    • @Prypak
      @Prypak Місяць тому

      I must add (even though it's obvious) that the power of 3^n and 7^n is gonna be the same, hence why you'll get either -2 or 2 when adding

  • @Ciruzzo_01
    @Ciruzzo_01 Місяць тому +1

    Great video prime newtons!

  • @tejpatel3763
    @tejpatel3763 Місяць тому

    Goes to show there is always more than 1 way to prove a statement in mathematics.
    I have a request: by any chance are you able to talk about transfinite induction? I encountered it in my set theory class but apart from that, I don't really see anyone explaining it correctly. Thanks!

  • @anonymouscheesepie3768
    @anonymouscheesepie3768 Місяць тому

    Base case: let n = 1
    3 + 7 - 2 = 8 = 2*4, thus P(1) is true
    Now, assume that 4 divides 3^k + 7^k - 2 for some k, k ∈ Z+
    Thus, 3^k + 7^k - 2 = 4A for some A, A ∈ Z+
    Now sub k + 1 in for n. Our statement becomes
    P(k+1) = 3^(k+1) + 7^(k+1) - 2
    P(k+1) = 3*3^k + 7*7^k - 2
    We assumed that 3^k + 7^k - 2 = 4A. Hence,
    3^k = 4A - 7^k + 2.
    ->
    P(k+1) = 3(4A - 7^k +2) + 7*7^k - 2
    P(k+1) = 12A - 3*7^k + 7*7^k + 4
    P(k+1) = 12A + (7 - 3)7^k + 4
    P(k+1) = 12A + 4*7^k + 4
    P(k+1) = 4(3A + 7^k + 1)
    If k is an integer, then 7^k is also an integer. As such, the sum 3A + 7^k + 1 is demonstrably an integer, showing that since P(k+1) is four times said sum, P(k+1) is also divisible by four. Thus, since our base case is true, and since P(k) -> P(k+1) is true, P(n) must be true for all n, n ∈ Z+.

  • @gp-ht7ug
    @gp-ht7ug Місяць тому

    One question: at 3:17 m belongs to Z which includes also the negative number. Why Z and not just N?

    • @justnowi8967
      @justnowi8967 Місяць тому

      Because we're writing a general statement that a number is divisible by 4.
      For example, in some definitions you may consider -4 to be divisible by 4, so m=-1,
      If that's unacceptable, still
      0 is divisible by 4, and 0 isn't included in natural numbers, though whole numbers will work.
      in this case 3^k+7^k-2>0 ,
      Still we're writing in general so we use Z.

  • @TurquoizeGoldscraper
    @TurquoizeGoldscraper Місяць тому

    Started with f(n) = 3^n + 7^n - 2
    Case n = 0:
    f(0) = 3^0 + 7^0 - 2
    f(0) = 1 + 1 - 2
    f(0) = 0
    f(0) = 0 (mod 4)
    Therefore f(0) is divisible by 4.
    Assume f(n) is divisible by for for all 0

  • @FranciscoMNeto
    @FranciscoMNeto Місяць тому

    Nice video!
    But when using modular arithmetic wouldn't you need to actually prove that 3^n and 7^n are congruent to 1 (mod 4) for all even n and to 3 (mod 4) for all odd n?

    • @bjornfeuerbacher5514
      @bjornfeuerbacher5514 Місяць тому +1

      "wouldn't you need to actually prove that 3^n and 7^n are all divisible by 4 as well?"
      Err - no? They aren't divisible by 4 on their own. Only the sum 3^n + 7^n - 2 is divisible by 4.

    • @FranciscoMNeto
      @FranciscoMNeto Місяць тому +1

      @@bjornfeuerbacher5514 I misspoke. I meant "prove that 3^n and 7^n are 1 (mod 4) for all even n and 3 (mod 4) for all odd n".
      Edited original comment to correct.

  • @spdas5942
    @spdas5942 Місяць тому +2

    Very nice application of method of mathematical induction. ❤from India.

  • @azaz700
    @azaz700 Місяць тому

    Very nice.
    Shouldn't the question also indicate that the result of the division should be a whole number for clarity's sake?

  • @misterj.a91
    @misterj.a91 Місяць тому

    3^n + 7^n - 2 = 3^n + (3+4)^n -2
    (3+4)^n = 3^n + 4^n + some combination of strictly positives powers of 3 and 4.
    4^n is divisible by 4 and any combinations and 3 and 4 contains 4 meaning it's divisible by 4.
    So
    3^n + 7^n - 2 = 3^n + 3^n - 2 mod 4
    = 2 * (3^n -1) mod 4
    (3^n is odd so (3^n -1) is even and could thus be replaced by 2k)
    =4k mod 4
    =0 mod 4

  • @surendrakverma555
    @surendrakverma555 Місяць тому

    Thanks Sir

  • @arthurkassis
    @arthurkassis Місяць тому

    I tried using mathmatical induction 'cause I'm more used to it
    first try the first natural number, n=0
    1+1-2=0 which is divisible by 4
    now, suppose that it works for a natural m, so
    3^m + 7^m - 2 = 4k, k is a nutural number
    then, let's see if it works to n=m+1
    3^m+1 + 7^m+1 -2
    = 3.3ˆm + 7.7ˆm -2
    =3.3ˆm + (3+4)7ˆm -2
    =3( 7ˆm + 3ˆm) + 4.7ˆm -2
    = 3.4k+6 + 4.7ˆm -2
    =4(3k+1+7ˆm) which is obviously divisible by 4

  • @sinekavi
    @sinekavi Місяць тому

    Can we do it by binomial theorem of (1+x)^n manipulation?

  • @pijanV2
    @pijanV2 Місяць тому +1

    you forgot to mention that n is an integer

  • @igorrromanov
    @igorrromanov Місяць тому

    3^x + 5^x will be divisible by 4 for any odd values of x.
    5^x + 7^x will be divisible by 4 for any odd values of x
    5^x + 9^x will be NOT divisible by 4 for any value of x.
    It is more generalized problem.

  • @kennethgee2004
    @kennethgee2004 Місяць тому

    divisibility is better understood as = 0 mod divisor. so we will look at 0 mod 4 for this equation. technically this is also not equal but congruent, but i do not know the how to write the equal sign with three lines in a browser.

  • @lepton56
    @lepton56 Місяць тому +2

    Wonderful presentation. Loved every moment of this.

  • @jayofray
    @jayofray Місяць тому

    3 and 7 each equal -1 mod 4, so 3^n and 7^n are (-1)^n mod 4. For n even, get 1+1-2=0 mod4, for n odd get -1-1-2= -4= 0 mod 4.

  • @simplicity530
    @simplicity530 Місяць тому

    4 l (3^n + 7^n - 2)
    4 l (3^n + (3 + 4)^n - 2)
    (3 + 4)^n = nC0 4^n × 3^0 + nC1 4^(n-1) × 3^1 + ... + nCn 4^0 × 3^n
    Only the last term, which is 3^n, isn't divisible by 4.
    4 l 3^n + 3^n - 2 = 4 l 2(3^n - 1)
    3^n - 1 is divisible by 2. The reason why this is always true is because 3 is an odd integer and an odd integer that raised to any integer power gives an odd integer. And an odd integer minus an odd integer gives an even integer.
    Let 3^n - 1 = 2M
    4 l 2 × 2M
    4 l 4M

  • @russellsharpe288
    @russellsharpe288 Місяць тому

    For even n=2m, 3^n + 7^n = 3^2m + 7^2m = (3^2)^m + (7^2)^m = 9^m + 49^m = 1^m + 1^m mod 4 ie 1+1=2 mod 4.
    For odd n = 2m+1, 3^n + 7^n = 3.1^m + 7.1^m mod 4 ie 3+7 = 10 = 2 mod 4.

  • @sr6424
    @sr6424 Місяць тому +1

    I believe I have a better method. Write 3 as 4-1) and 7 as(8-1). Expand the number of- (4-1)^2 is 4^2-4*2+1 etc. similar with 8-1. Every term is divisible by 4 except the 1 on the end. If n is positive it will be plus 1 if negative it will be minus. Add these to the minus 2 and the answer will be 0 or mins 4. Obviously both are divisible by 4.

  • @Anmol_Sinha
    @Anmol_Sinha Місяць тому

    We can also use binomial theorem by taking 2 cases - n is even and odd

    • @nimaalz4513
      @nimaalz4513 Місяць тому

      whats binomial theorem ?

    • @Anmol_Sinha
      @Anmol_Sinha Місяць тому

      @nimaalz4513 its like an expansion technique ig. 7^n = (3+4)^n = 3^n + 4×n×3^n-1... till.4^n.
      You can search it up, i don't think I will be able to do a good job in explaining it

  • @sobolzeev
    @sobolzeev Місяць тому

    The point is, this expression is divisible by 8, not just by 4. Care to prove? It seems easy, with the use of residues. Consider first n=2k (even case). Then our expression is
    9^k + 49^k -2. Since 9=49=1(mod 8), the result is immediate. Now let n=2k+1 (odd). We have
    3×9^k + 7×49^k -2
    = 3+7-2 = 0 (mod 8). So the result is immediate as well.
    I wonder if it is possible to do it without the residues.

    • @boguslawszostak1784
      @boguslawszostak1784 Місяць тому

      In the expansion of the binomial (a+b)^n, all terms are divisible by a, except possibly the last term.
      Instead of 9^k mod  8, take (8+1)^k and expand it using the binomial theorem.

  • @pijanV2
    @pijanV2 Місяць тому

    can you prove that the 3^n power is always equal to 1 mod 4 and 3 mod 4 repeating ?

    • @PrimeNewtons
      @PrimeNewtons  Місяць тому

      It's easy to prove

    • @pijanV2
      @pijanV2 Місяць тому

      @@PrimeNewtons how?

    • @PrimeNewtons
      @PrimeNewtons  Місяць тому

      Just prove for n is even by mathematical induction. Then prove for n is odd by multiplying by 3.

  • @rpggamers7867
    @rpggamers7867 Місяць тому

    I believe that solving this with Chinese remainder theorem may lead to a quicker solution, though your approach is more witty

  • @hrishikeshkulkarni8955
    @hrishikeshkulkarni8955 Місяць тому

    3^n + 7^n - 2 = (4 - 1)^n + (8 - 1)^n - 2 = > expands to powers of 4 using binomial theorem because the 1 + 1 - 2 cancels or becomes 1 + 1 + 2 = 4....

  • @沈博智-x5y
    @沈博智-x5y Місяць тому +1

    11:38
    you meant to write 7^n here since 7 = 3 (mod 4) only

  • @mircoceccarelli6689
    @mircoceccarelli6689 Місяць тому

    👍

  • @davidhuang8252
    @davidhuang8252 22 дні тому

    (7^n/2+1)(7^n/2-1)+(3^n/2+1)(3^n/2-1) each term can be divisible by 2

  • @RyanLucroy
    @RyanLucroy Місяць тому

    Well, my induction looked like this:
    3*3^n + 7*7^n -2 =
    (4-1)*3^n + (8-1)*7^n-2+4-4 =
    4*3^n + 8*7^n - 4 - (3^n + 7^n -2)
    Since every part is divisible by 4, the whole thing is divisible by 4.

  • @JohnPaul-rb7uk
    @JohnPaul-rb7uk 28 днів тому

    Wow

  • @thomazsoares1316
    @thomazsoares1316 Місяць тому

    Make a video proofing n⁴-6n³+23n²-18n + 24 is divisible by 24 for every n ≥ 0

  • @jcfgykjtdk
    @jcfgykjtdk Місяць тому +4

    Just do: 3 = 4-1 and 7 = 8-1. (4-1)^n + (8-1)^n = 2 mod 4 trivial

    • @PrimeNewtons
      @PrimeNewtons  Місяць тому +2

      You don't want to make a UA-cam video this way. You wouldn't be helping many people. Your skill level is for a select group.

    • @sobolzeev
      @sobolzeev Місяць тому +1

      ​@@PrimeNewtonsHowever, you can get a better result this way. Actually, your expression is divisible by 8, not just 4.

    • @boguslawszostak1784
      @boguslawszostak1784 Місяць тому

      @@PrimeNewtons My thoughts followed this course.
      (-1)mod 4= 3 mod 4=7 mod4
      so we can substitute
      ((-1)mod 4)^n+ ((-1) mod 4)^n -2 mod4=0
      (-1)^n+(-1)^n=2mod4
      For even n we get
      1+1=2
      for odd n we get
      (-1-1)mod4=2
      Sir, forgive me please
      I would gladly create films as beautiful as yours, but I lack the ability.

    • @boguslawszostak1784
      @boguslawszostak1784 Місяць тому

      @@sobolzeev (-1)^n mod 4 is easy to calculate , and that is the idea behind this solution.

    • @sobolzeev
      @sobolzeev Місяць тому

      @@boguslawszostak1784 Try n=2k, 2k+1. 3²=7²=1 mod (8)

  • @eloncole5702
    @eloncole5702 Місяць тому

    Allow me to spill your secret of catching viewers from one of your subscriber 😅 but it's all because of your captivating voice 😂❤ because once we open your videos it never gets bore instead it becomes more interesting anyway keep going and hoping to see you reaching 1m subscriber ❤️. POV : Bring some challenging topics 👍

  • @ccdsah
    @ccdsah Місяць тому +2

    3=4-1, 7=8-1, just expand the terms and you prove it, easy

  • @brnico6184
    @brnico6184 Місяць тому

    Incredible!

  • @katlegotshabalala6250
    @katlegotshabalala6250 Місяць тому

    If 3^n +7^n -2 =4m
    Then 2= 3^k+7^k-4m
    From this hypothesis I have 3^k+1+7^k+1 -(3^k+7^k-4m)
    Which simplifies to 4m +2(3^k +3*7^k)
    3^K +3*7^K IS divisible by 4
    Great sir!!!

    • @robertveith6383
      @robertveith6383 Місяць тому

      No, your third line simplifies to 4m + 2. The (k + 1) exponents must be inside grouping symbols.

  • @mazenzidieh
    @mazenzidieh Місяць тому

    Nice

  • @rpggamers7867
    @rpggamers7867 Місяць тому

    Glad I'm first here

  • @holyshit922
    @holyshit922 Місяць тому +1

    n=0
    1+1-2 = 0 , 4|0
    Assume that statement is true for n=k >= 0
    Check if assumption implies true statement for successor of k
    3^{k+1}+7^{k+1} - 2 = 3*3^{k}+7*7^{k} - 2
    3^{k+1}+7^{k+1} - 2 = 3*(3^{k} + 7^{k} - 2) + 4*7^{k}+4
    3^{k+1}+7^{k+1} - 2 = 3*(3^{k} + 7^{k} - 2) + 4(7^{k} + 1)
    3^{k} + 7^{k} - 2 is divisible by 4 by assumption , 4(7^{k} + 1) is divisible by 4 because 7^{k} + 1 is natural number and is multiplied by 4

  • @benshapiro8506
    @benshapiro8506 Місяць тому

    7 should b 7^n at bottom of left panel in method 2.
    also can show if x=y mod w then x^n = y^n mod w. otherwise, good presentation, PN.

  • @crossflux971
    @crossflux971 Місяць тому

    For the Mathematical Induction Proof, I understand that assuming 3^k + 7^k - 2 = 4m is crucial, but why can we make that assumption? If we were proving by contradiction, I get why we could make a claim and then prove it doesn't hold true.

    • @jonathanwarwick1117
      @jonathanwarwick1117 Місяць тому +1

      Because we want to show that if it is true for n=k then it is also true for n=k+1. Having done this, and since we have shown it is true for n=0 , we have a chain of reasoning that shows the statement’s truth for all values of n.

    • @ZachCameronWryNose
      @ZachCameronWryNose Місяць тому

      Making an assumption is always a valid move in the game of proofs. Every proof starts with some set of assumptions. If you make an assumption during a proof, you are starting a new subproof within your proof where you are operating under your initial assumptions as well as the new assumption you just introduced.
      For induction, your goal is to show that some statement P(n) is true over some range of integer values for n. In this video, the statement P(n) is defined as: there exists an integer m such that 3^n + 7^n - 2 = 4m. We want to prove that P(n) is true for all n≥0. The base case P(0) is shown to be true. A couple more cases, P(1) and P(2), are checked and confirmed as well. We wonder if this chain of true statements continues. We can't exhaustively check all of the infinite cases though. We'll have to stop up to some case P(k). Can we prove that the next case P(k+1) must be true regardless of what value of k we stopped at? To do so, you assume P(k) and show that P(k+1) follows. If you do so, you've proved that if P(k) is true, then P(k+1) is true for arbitrary k≥0. So even though we only checked cases up to P(2), we can be sure that P(3) also holds, and so we can be sure that P(4) holds, etc.

    • @crossflux971
      @crossflux971 Місяць тому

      @@jonathanwarwick1117 That's just it though - "IF" it is true for n=k. We haven't actually proved that. We can show n=k+1 just fine if we assume it is true for n=k. But how do we prove n=k is true?

    • @jonathanwarwick1117
      @jonathanwarwick1117 Місяць тому

      @@crossflux971 But we have proved it’s true for the base case n=0. Therefore it is true for n=1. Since it’s true for n=1, it must be true for n=2 etc.

    • @crossflux971
      @crossflux971 Місяць тому

      @jonathanwarwick1117 I'm still not following. Why does the Base case of n=0 prove that n=1 also holds true?

  • @nedmerrill5705
    @nedmerrill5705 Місяць тому

    Don't you need to specify, up front, that n is in *Z* ?