Prove that the square of any integer has the form 4k or 4k +1 for some integer of k.?
by skill of the branch set of rules, any integer n could properly be written interior the variety 4q + r for some integers q and r, the place r = 0, a million, 2, or 3. Case a million: r = 0. n^2 = (4q + 0)^2 = 16q^2 = 4(4q^2). Case 2: r = a million. n^2 = (4q + a million)^2 = 16q^2 + 8q + a million = 4(4q^2 + 2q) + a million. Case 3: r = 2. n^2 = (4q + 2)^2 = 16q^2 + 16q + 4= 4(4q^2 + 4q + a million). Case 4: r = 3. n^2 = (4q + 3)^2 = 16q^2 + 24q + 9 = 4(4q^2 + 6q + 2) + a million. In precis, we see that n^2 is of the variety 4k or 4k + a million for some integer ok. consequently, there is not any appropriate sq. of the variety 4k + 3. i'm hoping this facilitates! -------------------- observe: in case you recognize modular arithmetic, then you definately can streamline this argument extensively. working mod 4, any integer n = 0, a million, 2, or 3 (mod 4). So, n^2 = 0^2, a million^2, 2^2, or 3^2 (mod 4). ==> n^2 = 0 or a million (mod 4). observe that n^2 by no skill equals 3 (mod 4), and we are completed.
If n is a perfect square, prove that n+2 is not a perfect square?
Suppose that n = b^2 and n + 2 = a^2 for some non-negative integers a and b. Then, b^2 + 2 = a^2 ==> a^2 - b^2 = 2 ==> (a+b)(a-b) = 2 Since a, b are non-negative integers by assumption, we must have a + b ≥ a - b (as this is equivalent to b ≥ 0). Since this is an equation in integers, we have only one possibility: a + b = 2 and a - b = 1. ==> a = 3/2 and b = 1/2. This contradicts the fact that a and b are integers. -------------------- I hope this helps!
Is there any integer n, such that n(n+2) is a perfect square? If not, how can you prove it?
Note that [math]n(n + 2) = (m - 1)(m + 1) = m^2 - 1[/math] where [math]m = n + 1[/math] (the value halfway between [math]n[/math] and [math]n + 2[/math]). Setting this equal to a square gives us the equation [math]m^2 - 1 = x^2[/math], which is to say [math]m^2 - x^2 = 1[/math], which is to say, [math](m + x)(m - x) = 1[/math].Solutions now correspond to factorizations of 1 where both factors have the same parity. But there are are only two factorizations of 1 at all: there’s [math]1 = 1 \times 1 [/math] and there’s [math]1 = -1 \times -1[/math].The former factorization corresponds to taking [math]m + x[/math] and [math]m - x[/math] to both be [math]1[/math], and thus to taking [math]m = 1[/math], [math]x = 0[/math], and [math]n = 0[/math].The latter factorization corresponds to taking [math]m + x[/math] and [math]m - x[/math] to both be [math]-1[/math], and thus to taking [math]m = -1[/math], [math]x = 0[/math], and [math]n = -2[/math].Thus, we find there are precisely two solutions: [math]n = 0[/math] and [math]n = -2[/math].
How to prove that here is no perfect square of the form 4k+3?
exciting question. start up by using changing n with (3m + a million). This basically potential we are pondering basically values of n which may well be written as 3m + a million: a million, 4, 7, 10, 13, etc. So the expression turns into 3 + 6(3m + a million) = 18m + 9 = 9 (2m + a million). Now be conscious that 9 is a ideal sq., so if (2m + a million) have been additionally a ideal sq. then so too might the product be a ideal sq.. yet 2m + a million covers all ordinary numbers so for any ordinary ideal sq. S, we can discover an m such that 2m + a million = S. because of the fact the sq. of any ordinary quantity is ordinary, there are a limiteless form of such ordinary ideal squares and then a limiteless form of values of n. in case you like you may placed this all jointly right into a recipe for n by using working backwards interior the path of the above logic. a million) %. a unusual quantity 2k + a million: {a million, 3, 5, 7, ...} 2) sq. it to make S: {a million, 9, 25, 40 9, ...} 3) compute m such that 2m + a million = S it is, m = (S - a million) / 2: {0, 4, 12, 24, ... } 4) compute n = 3m + a million: {a million, 13, 37, seventy 3, ...} Or as a single step: n = 3m + a million = 3 (S - a million) / 2 + a million = 3 ((2k + a million)^2 - a million) / 2 + a million n = 3 (4 ok^2 + 4k) / 2 + a million n = 6 ok (ok + a million) + a million This holds for any ok so there are a limiteless form of values computable by using the above recipe and each has the valuables that 3 + 6n is a ideal sq.! {9, 80 one, 225, 441, ...}
Discrete Math : prove that the square of any integer has the form 3k or 3k+1 for some integer k.?
You've got the right idea. Let's say we want to figure out what form n^2 has. Divide this question into 3 cases: case one: n = 3m, for some integer m n^2 = (3m)^2 = 3*(3m^2) say that k = 3m^2, then n^2 = 3k case two: n = 3m + 1 n^2 = (3m+1)^2 = 9m^2 + 3m + 1 = 3(3m^2 + m) + 1 say that k = 3m^2 + m, then n^2 = 3k + 1 case three: n = 3m + 2 n^2 = (3m + 2)^2 = 9m^2 + 6m + 4 = 3(3m^2 + 2m + 1) + 1 say that k = 3m^2 + 2m + 1, then n^2 = 3k + 1 Since every integer n must fall into one of these 3 cases, n^2 must have the form of either 3k or 3k+1. QED
How do I prove that 4mn-m-n is not a perfect square?
Let [math]m=-1[/math] and [math]n=-3[/math]. The expression has value [math]16[/math] which is a perfect square. Check for any restrictions imposed on the question.However, if we insist that m and n are positive integers, we can continue as follows:Let [math]4mn - m - n = {a^2}[/math]. Then[math]4(m - \frac{1}{4})(n - \frac{1}{4}) - \frac{1}{4} = {a^2}[/math]and so[math]16(m - \frac{1}{4})(n - \frac{1}{4}) - 1 = 4{a^2}[/math].And then[math](4m - 1)(4n - 1) = {(2a)^2} + 1[/math].Or rather [math]{(2a)^2} \equiv - 1\bmod p[/math].Now [math]4m-1[/math] and [math]4n-1[/math] are both of the form [math]4k+3[/math] and therefore each must have a prime factor of that form.[math] ((4\alpha+1)(4\beta+1)=4(4\alpha\beta+\alpha+\beta)+1).[/math]So such a prime divides [math](2a)^2+1[/math]. But all prime divisors of [math](2a)^2+1[/math] are of the form [math]4k+1.[/math]
If n is an odd integer, then show that n squared minus 1 is divisible by 8?
If n is odd, then n can be written in the form 2x + 1, where x is an integer.[math](2x + 1)^2 - 1 = (4x^2 + 4x + 1) - 1 = 4x^2 + 4x[/math]Factoring out the 4:[math]4(x^x + x) = 4(x)(x+1)[/math]x and x+1 cannot both be odd…one of them must be even, which means one of them has a factor of 2. And since 4(x)(x+1) also has a factor of 4, then the whole thing must be divisible by 8.
How do i prove that "if n is a perfect square, then n is not the product of 2 and an odd number?"?
"If n is a perfect square, then n is not the product of 2 and an odd number." Remember that an odd number can be expressed in the form n = 2k + 1 (for an integer k). Let's try proving this by contradiction. Assume that n IS the product of 2 and an odd number. That is, assume n = 2(2k + 1). Then n = 4k + 2 Since n is a perfect square, it follows that sqrt(n) is an integer. Therefore, sqrt(4k + 2) is an integer. Factoring out a 4, sqrt(4[k + 2/4]) Reducing sqrt(4[k + 1/2]) Splitting up the square root into 2, sqrt(4)sqrt(k + 1/2) 2sqrt(k + 1/2) This means k + (1/2) is a perfect square. This is impossible (the square root of non-integers are non-integers), and this is a contradiction.
Why are there no integers x, y such that [math]x^2+y^2=2015[/math]?
This is because the remainder when 2015 is divided by 4 is 3.The square of any even number is divisible by 4:[math](2k)^2 = 4\times k^2[/math].The square of any odd number gives remainder 1 when divided by 4:[math](2k+1)^2 = 4k^2 + 4k + 1 = 4\times(k^2+k) + 1[/math].Thus, the square of any integer gives the remainder 0 or 1 when divided by 4.Now, if you add together [math]x^2+y^2[/math], what can be the remainder? We know that [math]x^2[/math] gives the remainder 0 or 1, and we also know that [math]y^2[/math] gives the remainder 0 or 1. Thus, the sum [math]x^2 + y^2[/math] will give one of the following remainders: 0+0, 0+1, 1+0, or 1+1.Hence, when divided by 4, the sum of any two squares can give one of the remainders 0, 1, 2, but never 3.(There are some nice elementary results from number theory related to this question. For example, Lagrange's four-square theorem is, in some sense, a generalization of the question asked here.)Another - less elegant but still correct - way to solve this problem is by using brute force. As both [math]x^2[/math] and [math]y^2[/math] can never be negative, each of them has to have a value between 0 and 2015, inclusive. So, obviously, we just need to check whether there is a solution with [math]0\leq x,y\leq\sqrt{2015}[/math]. This leaves us with only finitely many cases to check. We check them (or write a program to check them for us) and we see that none of them work. We lack the insight from the previous solution but we got the job done.
How can we show that if n is an odd perfect number, then n = p(a^2), where p is prime?
Eulers proof about odd perfect numbers actually shows something a little stronger than the version you gave, but it can be cut down to size.Given a prime factorization of [math]n[/math] we can calculate [math]\sigma(n)[/math], the sum of the divisors of [math]n[/math], as a multiplicative function with:[math]\sigma(p^e) = p^e + p^{e-1} + ... + p + 1 = \frac{p^{e+1}-1}{p-1}[/math]Note that [math]\sigma(p^e)[/math] (for [math]p \geq 3[/math]) is odd when [math]e[/math] is even, because it's an sum of an odd number of odd terms. And [math]\sigma(p^e)[/math] is even when [math]e[/math] is odd, because it's a sum of an even number of odd terms.Let [math]n[/math] be an odd perfect number, so that [math]\sigma(n) = 2n[/math].If [math]n[/math]'s prime factorization has no odd exponents, then [math]\sigma(n)[/math] is odd (a product of only odd terms), so that can't happen.If [math]n[/math]'s prime factorization has two distinct odd powers, then [math]\sigma(n) = \sigma(p_1^e)*\sigma(p_2^f)*X[/math] where [math]e[/math] and [math]f[/math] are odd, and so both those terms of the product are even. But then [math]\sigma(n)[/math] is divisible by 4, which also contradicts our assumption that [math]n[/math] is an odd perfect number.Therefore, if [math]n[/math] is odd and perfect, it can only have one prime factor with an odd exponent.