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"reducemry...." LONG PAUSE
I love that he keeps these little mistakes in. Honestly one of my favorite parts of these vids.
We could reduce it mod 10 at the beginning and then the solution is very straight forward. Primes mod 10 can give you only 1, -1 and then 5 for p=5 and 4 for p=2.
c^10 = (c^5)^2 mod 10 is easy to calculate in your head as well, 2^5 is well known, 5^5 mod 10 = 5, the rest is nearly the same. Then there is only one combination, because 1 and -1 mod 10 combinations don't work.
20:32
Why is d² equivalent 1 (mod 5)? Yes, it's odd, but since we're not modding an even number, it could still be 2 or 3 (mod 5), making d² equivalent to 4 (mod 5).
Edit: ah, b²+1 can't be 4. Right!
You are right 👍
2:39: +4. inconsequential, as both 4 and -4 are even and ≡4(mod 8).
17:22: Primes squared other than 5 do not have to be ≡1(mod 5). Because 5 is odd, odd numbers can appear even modulo 5. For example, 3²=9≡4(mod 5). Easy to show that all primes whose decimal representation ends in 3 or 7 have squares congruent to 4 mod 5. But those cases are easy because you immediately get b²≡2(mod 5), a contradiction.
b^2 = 3 mod 5, not, 2 mod 5, sorry. But anyways you arrive at a contradiction too.
It seems like the solution barely ever used the fact that a,b,c,d are primes.
I think if we loosen the requirements a bit we can still use the same methods to get all of the solutions.
If I'm not mistaken, you've used the primality for three observations:
1. each of these numbers can be either odd, or the single number 2
2. If a ≥ 5, then a ≡ 1,5 (mod 6)
3. If b² ≡ 0 (mod 5) then b = 5, and if d² ≡ 0 (mod 7) then d = 7
(4. also being primes means our search for solutions was narrower)
Maybe there were more of these, I'll check it tomorrow.
For section 1., we can just decide that we look for numbers that are either odd or the number 2.
Sure, it's not very natural, primes makes sense immediately, this set is weird. But I think it could be interesting to see what extra solutions we find then.
Also, section 2 was also pretty important, so let's... decide we want a to come from the set {2, 3} ∪ {n | n≡1,4 (mod 5) }
Now, section 3. is admittedly pretty important, so it probably won't be that easy, but I'm sure we could general
You need primality. There are infinitely many solutions without it in the form (a, b, c, d) = (r, s, 2t+1, s+1) where s = (10^r - 2t - 2)/2.
@@aadfg0 you forgot to raise c to the 10th, it should be s = [10^r - (2t+1)^10 - 1]/2, but yeah I get what you're saying.
I did specify that if we limit b,d to be either 2 or odd we can still use many of the arguments, so your family of solutions only adds only a few solutions (d=b+1 → b=2, d=3)
@@aadfg0 ok, if we set d-b=2 then we still get an infinite family
(a,b,c,d) = (r, s, 2, s+2), with
s = (10^r - 2¹⁰)/4 + 1
with r large enough so that the term in parentheses is divisible by 8, to meet all the requirements I've set
(a,b,c,d)=(3,5,2,1) also work
10^a+b^2= 1000+25=1025
C^10+d^2= 1024 +1=1025
1 is not a prime
*Prime numbers a,b,c,d*
7²-5²=24, but also 5²-1²=24.
1 is not a prime
*Prime numbers a,b,c,d*
You have to check that (a, b, c, d) = (3, 7, 2, 5) is the unique solution when d is odd and a = 3. You calculated that b = 7 and d = 5 is a solution for b^2-d^2 = 24, but is it the only solution? Yes: all the consecutive squares are apart 2n+1 units (because (n+1)^2 - n^2 = 2n+1). b cannot be larger than 7 cause the following square is 15 units apart (64) and the previous squares are at least 13 units apart (36), so their difference would be 28, at least (the difference with previous ones would be even larger, that is 15 + Sum(from 6 to k) of 2n+1, for each 1
I handled it a little differently albeit it's still brute forcing a bunch of cases. First take mod(4). LHS is 0 or 1. RHS is (0 or 1) + (0 or 1). So either b, c, and d are all even or b is odd and only one of c/d is odd. Since we're dealing with primes, being even means equal to 2.
(Case 1 - b = c = d = 2): 10^a + 4 = 1024 + 4 i.e. no solution.
(Case 2 - b odd, c = 11, and d = 2): I don't care what 11^10 is per se. I'm only interested in the fact that the last two digits must be 01 (this is obvious from either Pascal's triangle or binomial expansion). If I take the expression mod(5) ==> 0 + b^2 = 1 + 4 = 0 i.e. b^2 is divisible by 5. So b has to be 5 since it's prime. So the LHS is of the form 10^a + 25 where a>=2 so it's some number ending in 25. However, the RHS is a number ending in 01 + 4 = 05. Thus we have a contradiction.
(Case 3 - b odd, c odd and 11, and d = 2): c^10 = 1 mod(11) by Fermat's Little Theorem. So taking everything mod(11) ==> (-1)^a + b^2 = 1 + 4 = 5. The quadratic residues mod(11) are 0, 1, 4, 9, 5, and 3. So if a were odd, b^2 would have to be 6, which can't happen. So a = 2. Putting this back in, we get 96 = c^10 - b^2 = (c^5 + b)(c^5 - b), which has no solution for the same reason stated in the video.
(Case 4 - b odd, c = 2, d odd): You get an expression of the form 1024 - 10^a = (b+d)(b-d) and the justification follows exactly the same as the video.
Hello, Dr. Penn! Could you please prove, using number theory, that there are precisely 10 prime numbers within a set of 1000 consecutive natural numbers?
I used python to find between 1777555665727758999 and 1777555665727759999 (gap of 1000), there are less than ten prime numbers. used prime counting function approximation (n/ln(n)) for this.
Within any set of 1000 consecutive natural numbers? Then this isn't true. Take, for example, 1001!+n where n goes from 2 to 1001. Then regardless of what n is, 1001!+n is divisible by n and so can't be prime.
Also if that was true, then we can construct the prime counting function p(1000n) = 10n, which is not asymptotic to 1000n/ln(1000n)
@@thesecretguy2197 Sorry I worded it wrong. I ment any consecutive set of natural numbers. And using python and prime counting function, I was able to find the 1000 numbers with 8 primes and moving them to left or right its possible to show its possible. was lookingfor elegant proof
@@taterpun6211 I used prime counting function with a gap of 1000 numbers and Between 1,777,555,665,727,758,999 and 1,777,555,665,727,759,999, there are less than ten prime numbers (I used Python). If we adjust the range a little it works and we have such range. wasn't sure how to do it without computing.
well first thing that came to my mind was the a and c term, as 2^10 is approximately 10^3, then if thats part of a solution then b^2>25, b>5, so i just checked the next prime 7 and got 5 for d. {a,b,c,d} = {3,7,2,5}
Fascinating! Scratchpad jottings in this investigation must be quite a few pages - it is only comment I feel motivated to append.
..an endless subject..number theory
Bravo!
I wish I was smarter to get all this !
It's not been smart, it's practice and learning, if you take your time you will also be able to do it.
3^3
Yes but how is it related?