Set p = k-1. Then, n 2^n = p(p+2). Clearly p is even as n is positive. Set p = 2q. Gives n 2^(n-2) = q(q+1). This leads to n and 2^(n-2) being two consecutive integers (due to odd-even combination). This in turn means that there are only two solutions: n-1 = 2^(n-2) and n+1 = 2^(n-2) leading to n = 2 and 3.
Very smart thinking. I didn't see that n*2^(n-2) =q(q+1) initially but we cannot remove other factors than 2 from the 2^(n-2) part of the product and if we do remove factors then the difference between n*f and 2^(n-2)/f will be 0 or a multiple of 2 leading to the consecutive statement. Really nice and elegant
@@KiLLJoYRU-vidI really like the simplicity of this proof but I think that you are right the last step isn't quite right. The conclusion is only roght if n is odd or 2^(n-2) is odd. I think the best way to finish the proof is to note that the even part of n*2^(n-2) must be at most 1 unit larger than the odd part. This even part is greater than or equal to 2^(n-2) and the odd part is smaller than or equal to n. Since 2^(n-2)>n+1 for n>4 (2^3>5+1), you just need to check n=1,2,3,4. And, of course we get the right result then
Sorry for not clarifying the last step, which I put in paranthesis for the sake of brevity. However, my thought process was nicely summarized in the comments by @alucs6362. The only thing I would add is this. * If one of them, n or 2^(n-2), is odd, we are good. * Consider n is even. Since 2^(n-2) has no odd factors, n has to be a multiple of an odd number greater than or equal to 3, i.e. n > 5. Now, n is greater than the odd number and so 2^(n-2) lesser than the even number, leads to 2^(n-2) < (n+1). This inturn gives n < 5, which is a contradiction. So, n can't be even and a multiple of an odd number. Brings back to n and 2^(n-2) being consecutive.
Find all positive integers n such that n2^n+1 is a square. n2^n+1=k^2 2(2)^2+1=9=3^2 3(2)^3+1=25=5^2 k = -1, n = 0 k = 1, n = 0 k = ± 5, n = 3 k = ± 3, n = 2 Positive integers of n=2,3
Explanation of the Euclidean algorithm: If b = c (mod a), then gcd(a, b) = gcd(a, c). The reason for this is b = c (mod a) means b = c + ak for some k. Now, let d = gcd(a, b) and g = gcd(a, c). Since g divides a and c, it divides b. Thus, g divides a and b, and thus is a common factor of them. Thus, it divides the gcd of a and b, which is d. Now, rearranging b = c + ak to c = b - ak, it is clear to see that since d divides a and b, it also divides c. Thus, d divides a and c, and is thus a common factor of them, and thus divides their gcd, which is g. Thus, d and g divide each other, and so they are equal. So, in the problem, the reason gcd(x - 1, x + 1) = 2 is because first, we know x = 1 (mod x - 1), so that means x + 1 = 2 (mod x - 1) (just replace x with 1). Thus, gcd(x - 1, x + 1) = gcd(x - 1, 2). gcd(x - 1, 2) = 2 because we know x is odd, and so x - 1 is even, aka it has at least one factor of 2. In general, the Euclidean algorithm is nice for finding the gcd of large integers. Let's say we want the gcd of 78 and 96, we can take 96 mod 78 and get that we just need the gcd of 78 and 18, then we can take 78 mod 18 to get that we just need the gcd of 18 and 6, which is 6. Thus, the gcd of 78 and 96 is 6. In fact, you can even apply the Euclidean algorithm to polynomials. The gcd of f(x) and g(x) is the gcd of f(x) and g(x) (mod f(x)). You can obtain g(x) (mod f(x)) by polynomial long dividing g(x) by f(x) and taking the remainder. Make sure you long divide by the one of f(x) or g(x) with smaller (or equal) degree, because then the remainder will have reduced degree, which is what you want. After all, the purpose of an algorithm is to simplify a problem.
I just started looking at this, but I attacked it differently. I'm looking at trends when the statement is true. It is true when n=2 and n=3. This is as far as I've gotten in a few minutes.
Nice problem. If n2^n + 1 is a square then there is a positive integer x such that n2^n = x^2 - 1 = (x - 1)(x + 1). Taking both sides mod 2, it is clear that x must be odd, and thus x - 1 and x + 1 must both be even. Furthermore, the Euclidean algorithm gives us that gcd(x - 1, x + 1) = gcd(x - 1, 2) = 2 because x + 1 = 2 (mod x - 1). Thus, either x - 1 or x + 1 has a factor of 2^(n - 1). Note that for n >= 5, 2^(n - 1), which is the minimal size of the factor that does have 2^(n - 1) as a factor, is bigger than 2n, which is the maximal size of the factor that does not have 2^(n - 1) as a factor, and thus the larger of the two factors, x + 1, must be the one with 2^(n - 1) as a factor. Now notice that 2 = (x + 1) - (x - 1) >= 2^(n - 1) - (x - 1) >= 2^(n - 1) - 2n > 2. This is a contradiction, so we must have n < 5. n = 1 does not yield a solution because 3 is not a square n = 2 does yield a solution because 9 is a square n = 3 does yield a solution because 25 is a square n = 4 does not yield a solution because 65 is not a square Thus, the solutions are n = 2, 3
n*2^n+1 is an odd number, which it must be the square of an odd number, therefore we can write n*2^n+1=(2*k+1)^2. This leads to n*2^(n-2) = k*(k+1). We observe that k and k+1 are consecutive integers, which means they are coprime. |2^(n-2) - n| > 1 for n≥5. This means that n*2^(n-2) can't possibly be the product of two consecutive integers (for n≥5) since one of them must be at least 2^(n-2) (the other number has to be odd) and the other one can be at most n. That leaves only n ∈ {1,2,3,4} as possible candidates. n=1 doesn't work because 1*2^(1-2) is not an integer. n=2 works because 2*(2^(2-2)) = 2 can be factorized into consecutive integers 1*2 (k=1). Therefore 2*2^2+1 = 9 = (2*1+1)^2 n=3 works because 3*(2^(3-2)) = 6 can be factorized into consecutive integers 2*3 (k=2). Therefore 3*2^3+1 = 25 = (2*2+1)^2 n=4 doesn't work because 4*2^(4-2) = 16 can only be factorized into coprime factors as 1*16, which are not consecutive numbers. More values of n just for fun: n=5 -> 5*2^(5-2) = 40 --> coprime factorizations: 5*8, 1*40 n=6 -> 6*2^(6-2) = 96 --> coprime factorizations: 6*16, 3*32, 2*48, 1*96 n=7 -> 7*2^(7-2) = 224 --> coprime factorizations: 7*32, 1*224 n=8 -> 8*2^(8-2) = 512 --> coprime factorizations: 1*512 n=9 -> 9*2^(9-2) = 1143 --> coprime factorizations: 9*127, 3*381, 1*1143 ...
Method 1) (- x= 3) equation is given Multiplying both sides by (-1) -1*-x=-1*3 Then x=-3 or Method 2) Let the equation be (- x= 3) If we multiply both sides with "MINUS" sign -(- x)= -(3) Then x= -3. Which one is correct or both methods are correct . Please help 🙏🙏
n*2^n= k^2-1 n*2^n divisible by k+1 and k-1 n*2^n, n=k+1 or n=k-1, or n is combinable with 2^n. n is not combinable with 2^n, as no perfect square is adjacent to an integer exponent. 2^9+-1 cannot be a perfect square. or k+1,k-1 is divisible by something else. for n=k+1,n=k-1, 2 options: n-1 =k, or n+1=k 2^n+1=k or 2^n-1 =k n+1=2^n-1 n=2, 3=4-1, therefore 2 is a solution, no other solutions of this form, same reasoning. n-1=2^n+1 n=1, 0=3, therefore no solutions. for k+1 divislbe by something else: n= (k-1)/u, 2^n=(k+1)*u, and vice versa. 4 options, Most plausable case is the one stated: n*u+1=2^n/u-1 n=3, u+1=2^3/u-1 u=2, 3=3 n=3,u=2 is a solution. Check the other 3 for any other solutions, but i have finished the video so it is not neccesary. Wonderful as always, harder problem than i realized.
I have 2 questions : 1) why did we ignore n*2^n = (k-1)a2^(n-1) ? 2) n*2^n = (k-1)a2^(n-1) (and the other) It looks like it works but how to we prove it ?
