Can You Find The Easy Answer? Learn The Brilliant Trick!

Thanks to Ben Craft for sending me this problem and its solution! Find a value of x such that (3 + 4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^16 + 4^16)(3^32 + 4^32) = 4^x - 3^x. There's a really neat trick that simplifies the problem like magic! Thanks to Nar235 and Jonathan Love for finding a mistake in the wording! The earlier problem was solve for 4^x - 3^y, but that has infinitely many solutions.

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[tojarin]

I find that it is amazing how so many mathematical monstrosities can be simplified just by multiplying by one.

[rdpeterson]

I didn't see the neat trick. But I managed to solve it a different way. I started multiplying the terms out, and saw the pattern: 3^63 + 3^62x4 + 3^61x4^2 ... + 4^63. Then I reduced it to 4^63(1 + 3/4 + (3/4)^2 + ... + (3/4)^63). Then I was able to find the sum of that series using the sum of a geometric series (from your video awhile back). The sum of the series is 4 - (3^64/4^63), and it gives the solution x=64. Great problem!

[jean-baptiste]

For YouTube subtitle algorithm your name is Pressure Walker.

[samarendra]

I got the idea to multiply by 4-3 But then thought how can i multiply it without changing the value . Then I saw the video, and realised 4 -3 = 1. 😪

[santiagolopez-efhq]

Let x=2n. Then the right side of the equality can be written as 4^2n-3^2n=(4^n-3^n)(4^n+3^n). This implies that (4^n+3^n)=(3^32+4^32), and so n=32. Hence, x=64.

[jessstuart]

If you think of the exponents as bits of binary numbers, multiplying these binomial terms will generate all possible cross-terms who's exponents sum to the next higher power of 2 minus 1. All cross-terms are unique (cannot be combined through addition), so the coefficients of all cross-terms are equal to 1.

This expands to...

A^63*B^0 + A^62*B^1 + A^61*B^2 + ... + A^3*B^60 + A^2*B^61 + A^1*B^62 + A^0*B^63 = A^64-B^64

And the Generic Identity...

Sum(A^(n-i-1)*B^i, i, 0, n-1) = A^n - B^n where n is a power of 2.

[justpaulo]

Nice reasonable easy problem.
A: (4^64 - 3^64)

1) Start with (3+4). Multiply by (4 - 3)=1. You get (4^2 - 3^2).
2) Now substitute (3+4) by (4^2 - 3^2). When multiplied by (4^2 + 3^2) you get (4^4 - 3^4). Substitute (4^2 - 3^2)(4^2 + 3^2) by (4^4 - 3^4) and keep going in the same fashion, collapsing the all expression from left to right.
3) In the last step you'll get (4^32 - 3^32)(4^32 + 3^32) which equals (4^64 - 3^64).


Q: How do you prove there are infinite solution of the type (4^x - 3^y) ?

[LMANJINA]

I got x=64. I came to the conclusion before i was able to prove it (mainly due to it being a fairly arduous proof) but here was the mentality that i approached it with:
LHS = (4^x - 3^x)
= (4^x/2 + 3^x/2) (4^x/2 - 3^x/2)
= (4^x/2 + 3^x/2) (4^x/4 + 3x/4) (4^x/4 - 3^x/4)
= (4^x/2 + 3^x/2) (4^x/4 + 3^x/4) (4^x/8 + 3^x/8) (4^x/8 - 3^x/8)
= (4^x/2 + 3x/2) (4^x/4 + 3^x/4) (4^x/8 + 3^x/8) (4^x/16 + 3^x/16) (4^x/16 - 3^x/16)
= (4^x/2 + 3x/2) (4^x/4 + 3^x/4) (4^x/8 + 3^x/8) (4^x/16 + 3^x/16) (4^x/32 + 3^x/32) (4^x/32 - 3^x/32)
= (4^x/2 + 3x/2) (4^x/4 + 3^x/4) (4^x/8 + 3^x/8) (4^x/16 + 3^x/16) (4^x/32 + 3^x/32) (4^x/64 + 3^x/64) (4x/64 - 3x/64)

By inspection, it seems appropriate to substitute x = 64.

So when x = 64

=> (4^32 + 3^32) (4^16 + 3^16) (4^8 + 3^8) (4^4 + 3^4) (4^2 + 3^2) (4^1 + 3^1)(4 - 3)
=> (4^32 + 3^32) (4^16 + 3^16) (4^8 + 3^8) (4^4 + 3^4) (4^2 + 3^2) (4 + 3)
=> LHS

Quod erat demonstrandum.

:. x = 64

[Edit: i accidentally added an extra term in the second last line (4^0 - 3^0) which was not meant to be there, i got a bitt carried away writing the whole thing, it was supposed to be (4 - 3). Mathematically that term would never exist in the series, it would tend to that term but never reach it, so i thought i’d fix it].

[lightdark]

My guess was 64 immediately. It added to 63, but it was subtraction instead of addition so one needed to be added.

[SomeGuyx]

I immediately guessed 64, then did out the math and found it was correct, although I didn't see the exact trick right away. Pretty neat now that domino effect worked.
(3:13) "And once more... DECEARING EGG!" ...Oh, wait. That's a different video.

[thecrouchingdragon]

Glad to see so many people have worked it out without being intimidated by its length, it's an easy one though.

[jagmarz]

This is fantastic! What an amazing solution! I got the answer by noticing a pattern, but this was super elegant!

[willbishop]

Great problem! I used a slightly different trick which isn't quite as clever. We have a general formula that (a^n - b^n) / (a - b) = a^(n-1) + a^(n-2)*b + a^(n-3)*b^2 + ... + a*b^(n-2) + b^(n-1). Applying this to the problem, (4^x - 3^x) / (4 - 3) = 4^(x-1) + 4^(x-2)*3 + ... + 4*3^(x-2) + 3^(x-1). The highest exponent in that expression has to be 1+2+4+8+16+32 = 63, so x-1 = 63 and x = 64.

[cameronminty]

The way I solved this problem was discovering that (3+4)(3^2+4^2)...(3^32)(4^32) was also equal to 3^63+3^62•4+3^61•4^2+...+4^63 which is also equal to the summation of i=0 approaching 63 of 3^(63-i)•4^i, or 3^63•(4/3)^i. I then translated the entire summation using Gauss’ summation algorithm (k[SUM]i=0: [a(r^i)] = [a(r^[k+1]) - a]/[r-1], r#-1) to get the answer (3^63[4/3]^[63+1] - 3^63)/(4/3 - 1) = 3(3^63•4^64/3^64 - 3^63) = 3(4^64/3 - 3^63) = 4^64 -3^64. x = 64, and it was all proven discretely.

[seiger]

I solved it a different way (not quite as elegant):
= 4^x-3^x
multiplying the left side out
sum n=0 to 63 4^n*3^(63-n) = 4^x-3^x
divide both sides by 4^63
sum n=0 to 63 4^(n-63)*3^(63-n) = 4^x/4^63 - 3^x/4^63
simplifying
sum n=0 to 63 (3/4)^n = 4^x/4^63 - 3^x/4^63
truncated geometric series formula
(1-(3/4)^64)/(1-3/4) = 4^x/4^63 - 3^x/4^63
divide both sides by 4
(1-(3/4)^64) = 4^x/4^64 - 3^x/4^64
which has solution x=64

[thelatestartosrs]

if you can't find the solution, reduce the problem and try to write 4+3 as 4^n - 3^n and you'll find n=2 and then multiply by the next term and try to write 175 = 4^n - 3^n and you find n=4, then realise that:

(4^(2^n) - 3^(2^n))(4^(2^n) + 3^(2^n)) = 4^(2^(n+1))-3^(2^(n+1))

[donaldasayers]

Yeah got that! 64 was obviously the first value that had any chance of working.

[hpp]

That was a fun problem to solve. Thanks! I basically solved it, by employing the same trick as you did but, in a slightly different manner,
which might be a little bit more suitable if one wants to generalize the problem by not specifying the summands, i.e., the 3 and 4, in the
parentheses. So let us assume, that the problem was (a + b) * (a^2 + b^2) * (a^4 + b^4) * (a^8 + b^8) * (a^16 + b^16) * (a^32 + b^32) = ?.
First, I expanded the expression by (a^32 - b^32) / (a^32 - b^32), and then I noticed, due to (a^n + b^n) * (a^n - b^n) = a^(2n) - b^(2n),
that an avalanche of cancellation occurs, which leads - after following the cancellation steps - to (a^64 - b^64) / (a - b)!

In detail:
= (a + b) * (a^2 + b^2) * (a^4 + b^4) * (a^8 + b^8) * (a^16 + b^16) * (a^32 + b^32) * (a^32 - b^32) / (a^32 - b^32)
= (a + b) * (a^2 + b^2) * (a^4 + b^4) * (a^8 + b^8) * (a^16 + b^16) * (a^64 - b^64) / [ (a^16 - b^16) * (a^16 + b^16) ]
= (a + b) * (a^2 + b^2) * (a^4 + b^4) * (a^8 + b^8) * (a^64 - b^64) / (a^16 - b^16)
= (a + b) * (a^2 + b^2) * (a^4 + b^4) * (a^8 + b^8) * (a^64 - b^64) / [ (a^8 - b^8) * (a^8 + b^8) ]
= (a + b) * (a^2 + b^2) * (a^4 + b^4) * (a^64 - b^64) / (a^8 - b^8)
= (a + b) * (a^2 + b^2) * (a^4 + b^4) * (a^64 - b^64) / [ (a^4 - b^4) * (a^4 + b^4) ]
= (a + b) * (a^2 + b^2) * (a^64 - b^64) / (a^4 - b^4)
= (a + b) * (a^2 + b^2) * (a^64 - b^64) / [ (a^2 - b^2) * (a^2 + b^2) ]
= (a + b) * (a^64 - b^64) / (a^2 - b^2)
= (a + b) * (a^64 - b^64) / [ (a - b) * (a + b) ]
= (a^64 - b^64) / (a - b)

In our case a = 4 and b = 3. Thus, (a^64 - b^64) / (a - b) = (4^64 - 3^64) / (4 - 3) = 4^64 - 3^64.

[embedded_]

I just cancelled out left side 4^(1+2+4+8+16+32)=4^63 . Then I got (1+3/4)*(1+(3/4)^2)*... Then I got the pattern that it's sum of geometric progression: sum from n=0 to 63 (3/4)^n . And this is equal to So left side is Than we can expand the brackets and get 4^64-3^64

[BigDBrian]

now find the general formula for the given expression equal to 4^f(t) - 3^g(t) holds true for any value of t.
Alternatively set it equal to 4^x - 3^y, find y in terms of x.
Your choice which to use. Basically, create a formula which lets you easily find all the solutions.