Here's my take, no modular arithmatic, no binomial expansion, no number theory. Just pure brute force . Observe that 2^2013 + 13 is 2^2013 - 1 + 14. 14 is just 7x2 but 2^2013-1 can be expanded by using geometric series to 1 + 2 + 2^2 + 2^3 + ... + 2^2012, which is 111...111 in binary (2013 digits binary of all 1). Similarly, 7 is 2^3 -1, which can be expanded to 1 + 2 + 2^2 which is 111 in binary (3 digits of 1). Since 2013 is divisible by 3 (2+0+1+3 = 6), 111...111 must be divisible by 111. Why? Because the result is obviously 100100...1001 in binary (670 copies of100 followed by 1, 2011 digits total) which definitely is an integer. The cool thing when you brute force is that not only you've proven that 7 divides 2^2013+13 but you also got the quotient (although, it's in binary).

Before watching: 8-1=7 If 8^n-1 is divisible by 7, 8^(n+1)-1=7×8^n+8^n-1 is also divisible by 7. The rest is basic arithmetic (8=2^3, 2013 is divisible by 3, so is a power of 8, 2^2013+13 is 2×7 more than 2^2013-1).

(Before watching) My first thought is immediately Fermats Little Theorem, but I'm assuming we probably cant do that. Could be useful that 2³ = 1 mod 7.

Witty name for a channel 😊

hello, i think that ,in all integers( positives and negatives) , -2 is also a solution for n^2 + 1 divisible by n+1. is this correct?

Rewrite the requirement as A^2 + B^2 = 261 which is odd, so just try subtracting 0^2, 2^2, 4^2, ..., 16^2 from 261 to see if there are any squares. This just gives 261 = 6^2 + 15^2. Then (6, 15), (15, 6), (-6, -15) and (-15, -6) are the four solutions of the ordered pair (A, B).

It doesn't help much, but because 261 is odd, it can only be the sum of an odd and an even number. The square of an odd number is odd and the square of an even number is even. Hence we are looking for one odd and one even number.

Nice problem, but it fell out felicitiously when I tried it. It should be obvious immediately that the sum is simply (A+B)^2 - (2AB - 1) = A^2 + B^2 + 1 = 262 which gives A^2 + B^2 = 261. Once you have that, it's clear that A and B are no more than 16. That's a small enough range to use trial and error at that point. We are looking for a value of 261 - A^2 that is a perfect square. 261 - 16^2 = 5 doesn't work. However 261 - 15^2 = 261 - 225 = 36 = 6^2. That gives us the solution pairs (15, 6) and (-15, -6). It doesn't take many minutes to see that (261 - A^2) where A ∈ { 12, 13, 14 } do not produce any more perfect squares. So the solutions found are unique. (Edit: because 2 * 11^2 < 261).

I got accepted into there but didnt have passport ToT

Am I the only person who thought this was such a weirdly specific problem?! To the point of being contrived.

a+3b=24; a=0, b=8; a=3; b=7; a=6, b=6; a=9, b=5; 12, b=4; a=15, b=3; a=18, b=2; a=21, b=1; a=24, b=0 c follows from either equation.

Schmitt Land

108 mod 7 = 3 24 mod 7 = 3 108^n - 24^n = 3^n - 3^n (mod 7) = 0 (mod 7)

This is instructive but for this exemple we don’t need Muirhead inequality Juste use : a+b >= 2sqrt(ab)

1:12 in theory, you could do induction in both directions to prove the statement for all integers, rather than just the nonnegative ones.

Due to Euler's theorem with the totient function we know that for primes p any a with 1 < a < p we got a^2 = 0 mod p (because primes are always coprime with smaller positive integers). That means n^p = n mod p, which means n^p-n = 0 mod p, so we don't need to test 3 and 5. It also means for instance n^11 - n = 0 mod 11 or n^23 - n = 0 mod 23.

You demonstrated it was true for n = 1. Let's denote f(n) = n^5 / 5 + n^4 / 2 + n^3 / 3 - n / 30. So... f(n+1) = (n+1)^5 / 5 + (n+1)^4 / 2 + (n+1)^3 / 3 - (n+1)/30. But, what I'm actually going to be interested in is f(n + 1) - f(n), so... f(n+1) - f(n) = [(n+1)^5 - n^5] / 5 + [(n+1)^4 - n^4] / 2 + [(n+1)^3 - n^3] / 3 - [(n+1) - n]/30 f(n+1) - f(n) = [5n^4 + 10n^3 + 10n^2 + 5n + 1] / 5 + [4n^3 + 6n^2 + 4n + 1] / 2 + [3n^2 + 3n + 1] / 3 - 1/30 f(n+1) - f(n) = n^4 + 2n^3 + 2n^2 + n + 1/5 + 2n^3 + 3n^2 + 2n + 1/2 + n^2 + n + 1/3 - 1/30 f(n+1) - f(n) = n^4 + 4n^3 + 7n^2 + 4n + 1 So, if n is an integer, then f(n+1) - f(n) is an integer. And therefore, if n and f(n) are both integers, then f(n+1) = f(n) + [f(n-1) - f(n)] is the sum of two integers, so it's also an integer. So, since 1 and f(1) are both integers, then f(2) is an integer. And since 2 and f(2) are both integers, then f(3) is an integer. And since 3 and f(3) are both integers, then f(4) is an integer. This process repeats forever, so for any positive integer n, f(n) is an integer. But the problem asked for all n... Well, if n is an integer, and f(n+1) is an integer, then f(n+1) - [f(n+1) - f(n)] is the difference of two integers, so it's also an integer. And 0 and f(1) are both integers, so f(0) is also an integer. And -1 and f(0) are both integers, so f(-1) is also an integer. This process also repeats forever, so for 0 and any negative integer n, f(n) is an integer. So, overall, for any positive integer n, f(n) is an integer, which is what we wanted to prove.

387=1×387=3×129=9×43 LCM=388 or 390 or 396 Actual answers are with GCD 1 or 3 or 9 (1,388),(4,97),(3,390),(6,195),(5,78),(30,39),(9,396) and (36,99)

Correction in your reductions: 1) 342/117 reduces by 9/9 (not 13) to 38/13 (which is what you said, but wrote 38/3) 2) 338/117 reduces by 13/13 to 26/9 (not 27/9, which would reduce to 3 and be higher than the other root)

Hi

They are 2,3,5,7,17.

The answer is -4413+3080i. I shall compare this problem to the other problem!!!

Bro, just delta=b^2-4ac <0

If an is a square and bn is a cube then n=a^3×b^2 Here a=6,b=7 n=6^3×7^2.

That is literally one of the best examples of "easier than it looks"!!!. I better use that as practice!!!

A little less messy is to compute x^2 + y^2 and x^3 + y^3. Multiply the two and bingo/

There is a standard idea: find one rational solution and intersect a given 2-nd order curve with any line with rational coefficents goes through the first solution. Second intersection will be rational. If we start with a trivial solution (2/73,0) and the line y=x-2/73 we will get your solution (137;9999/73).

Given circle has a few integer points but infinitely many rational points. If scholar believes in teacher's good will he will look for whole points. If not than he had to find rational root of 4-th degree equation with a rather big coefficients via a well-known theorem about rational roots.

if f(n) = x^n +y^n, simple manipulation shows f(n)*f(1) = f(n+1)+xy*f(n-1) or f(n+1) = f(n)*f(1) - xy*f(n-1) with the givens and the offhand observation that f(0) = 2, we can just chase the recurrence for 4 very easy steps. f(2) = 3*3 - 7*2 = -5 f(3) = 3*-5 - 7*3 = -36 f(4) = 3*-36 -7*-5 = -73 f(5) = 3*-73 -7*-36 = 33

Same idea, slightly different approach ... (x+y)⁵=243=x⁵+5x⁴y+10x³y²+10x²y³+5xy⁴+y⁵ =x⁵+y⁵+5xy(x³+2x²y+2xy²+y³) =x⁵+y⁵+35(x³+3x²y+3xy²+y³-(x²y+xy²)) =x⁵+y⁵+35((x+y)³-xy(x+y)) =x⁵+y⁵+35(27-21) =x⁵+y⁵+210 x⁵+y⁵=33

Here is something that is almost purely computational (20^5 + 70^5) / (20^4 - 5460000 + 70^4) = (10^5 * (2^5 + 7^5)) / (10^4 * (7^4 - 546 + 2^4)) = 10 * (32 + 7 * (50 - 1)^2) / (49^2 - 2*49*4 + 4^2 + 2*49*4 - 546) = 10 * (32 + 7 * (2500 - 100 + 1)) / ((49 - 4)^2 + 8 * (50 - 1) -546) = 10 * (32 + 7 * 2401) / (45^2 + 400 - 8 - 546) = 10 * (32 + 16807) / (25 * 81 - 154) = 10 * 16839 / (8100 / 4 - 154) = 10 * 3 * 5613 / (2025 - 154) = 30 * 3 * 1871 / 1871 = 90

(x+2)/5 + 5x - 26x = 2/5

how did you come to conclusion that 3n - (x+y+z)?

As f has a max power of 4, f(f) can only have a max power of 4*4=16

Alternatively, as the largest exponent e where the coefficient x^e of f(f(x)) (where f(x) = x^2 + x^4) is nonzero is the same as the largest exponent e where the coefficient of x^e of g(g(x)) where g(x) = x^4 so g(g(x)) = (x^4)^4 = x^16, the coefficient of x^18 in f(f(x)) is 0.

This is one of those cases where we say the emperor is naked

GREAT VIDEO! Liked and subscribed ❤

I'm a secondary Maths teacher in the UK, loved this! Thanks!

31459 is the largest 5 digit fibbish number i could find

31

Great puzzle.

The result is indeed 38, and to prove it all you need to do is show that 2^122 + 1 is a multiple of 2^61 + 2^31 + 1. A quick way to do it is observe that (x^61 + x^31 + 1)(x^61 - x^31 + 1) = x^122 - x^62 + 2 x^61 + 1, then replace x with 2 and get (2^61 + 2^31 + 1)(2^61 - 2^31 + 1) = 2^122 + 1, voilà!

Perhaps a silly way to do it, lol, but write them out in binary...you get 2^122 + 32 + 7=2^122 + 2^5 + 2^2 + 2 + 1, which, in binary is 100...000100111 for the bottom, etc, and the bottom becomes 100...1....01, etc...with that "1"in the middle the 32nd digit from the right...and just do long division, lol...long division in binary...I keep getting mixed up, lol, but it should give the answer...just be mindful of the missing digits, one can't write them all out...maybe there's a much easier way, lol...I tried just guessing that the largest multiple smaller than the numerator (of the denominator) would be 2^60+2^59...+ 2^2 + 2^1 + 2^0, etc, all 1's in binary...you would multiply that by the denominator and then just subtract, I went back to binary thinking that was even more cumbersome, lol...

I tried to solve this by long division, as though it was algebraic long division, and it works. The denominator goes into the numerator 2^61 - 2^31 + 2 - 1 with a remainder of 38.

You made it look so easy. I am pretty sure i would have wasted at least half to one hour on this.

Nice but i dont get phi function what it is doing u need to explained it at first u only tell it when prime

Maybe look at prime signatures. Have you tried n = 20?

I like the proof. You could have shortened it by looking at sqrt(720)=26.8...<(2p+1) => p=13.

Correct to k=1 only. Start by assuming only two factors a,b, both are odd. Sum of factors S=(1+a+a²+a³)(1+b+b²+b³) but the bracketed expression are just geometric series so S=((a⁴-1)/(a-1))*((b⁴-1)/(b-1)) sub in minimum odd prime numbers 3, 5 (ignoring 1 because it is obviously not of this form) S=(80/2)*(624/4) S=40*156 >400 So there cannot be two odd prime factors, there must be less than two, i.e. only 1 looking at the numbers for 5, try 7 S=(7⁴-1)/(7-1) S=2400/6=400 so 7 is the odd factor. Given that even factors play no part and we can only have one odd factor, the other factor must be a power of 2. Given the requirement that it is a perfect cube then the power must be 0, 3, 6, 9, 12, ... So n=(2³)^k*7³ for k=0, 1, 2, 3, 4, ... For k= 0 or k>1 the number of factors does no match, so k=1 gives the only solution.

2.14*200 = 428 so i guess the answer is 429. I liked you're explanation of Cauch Shwartz.