Let n, m be an element of Z+ and a,b be elements of Z.  Suppose that a=b(modn) and a =b(modm) and that (n,m)=1
337664
Z+ means positive integers.\n\nProof.\n(a - b) is divisible by n.\n(a - b) = pn for some positive integer p.\n\n(a - b) is divisible by m.\n(a - b) = qm for some positive integer q.\n\nSince both m and n divide (a - b) and have no factor in common (from (n,m)=1 ), it means that (a - b) is a multiple of (mn) for m & n each contribute DISTINCT FACTORS into the composition of (a - b). This means (a -b) is divisible by m and n "SIMULTANEOUSLY".In other words, (mn) divides (a - b).\n\nThat is, (a - b) = 0 ( mod mn )\nor a=b(mod nm)\n\nQED\n\nRegards
769753
Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
484601
The previous poster didn't see the whole problem. This is a question about modular arithmetic, not standard algebra. The equations are to be done modulo n and the solutions are to be integers.\n\nBy adding 3 times the first to the second, we need 5x=8. By subtracting twice the first from the second, we get -5y=-2, so 5y=2. So all that is required is that 5 be invertible modulo n, i.e., that 5 does not divide n. For example, if n=4, x=0 and y=2 works.
487838
1) Assuming an aritmetic sequence\n\nbn = b1 + (n-1)d \nletting b3=a1 = 20, then b7=a5 = 32, and b25 = b23\na5 = a1 + (5-1)d\n32 = 20 + (5-1)d\n32 = 20 + 4d\n12 = 4d\n3 = d\n\nb25 = a23 = 20 + (23-1)*3\nb25 = 20 + 22*3\nb25 = 20 + 66\nb25 = 86\n=====================================================\n2)\na75 = a1 + (75-1)d; d=-3\n\na75 = 5 + (75-1)*-3\na75 = 5 + 74*-3\na75 = 5 + -222\na75 = -217\n\nSn = n/2(a1+an)\nS75 = 75/2(5 + -217)\nS75 = 75/2(-212)\nS75 = -7950\n===================================================\n3) ????\n\na5 = 5*5 - 4 = 21\na15 = 5*15 - 4 = 71\na25 = 5*25 - 4 = 121\n\nOR MAYBE\n\na1 = 5*1 - 4\na1 = 5 - 4\na1 = 1\na30 = 5*30 - 4\na30 = 150 - 4\na30 = 146\n\nSn = n/2(a1+an)\nS30 = 30/2(1 + 146)\nS30 = 30/2(147)\nS30 = 2205\n================================================\n4)\n\nrth term is nCr-1m^n-(r-1)k^r-1\n\n5th term =\n4C4m^4-(5-1)k^5-1\n4C4 = 1\n\n5th term = k^4
460848
no there doesnt. If you solve the congruences one at a time and use your solution for the previous congruence in the next one, you will eventually get the colution 1 congruent to 0 (mod7) which is obviously impossible
942422
You can start from the following identity for set differences which holds true for any sets X, A, and B\n\nX − (A ∪ B)  =  (X − A) ∩ (X − B)\n\nNow, if B is a subset of A, then A ∪ B = A\n\nthe identity above becomes\n\nX − A =  (X − A) ∩ (X − B)\n\nFor this to hold true, X − B must be a superset of (X − A), or in other words, (X − A) must be a subset of X − B
758829
a = 6 and b = 8\n6^2 + 8^2 = 36+64 = 100\n\nor a = 8 and b = 6\n8^2 + 6^2 = 64+36 = 100\n\nor a = 10 and b = 0\n10^2 + 0^2 = 100+0 = 100\n\nor a = 0 and b = 10\n0^2 + 10^2 = 0+100 = 100\n\nWithout any restrictions on a and b, there are infinitely many answers.
164974
The non-brute-force solution is to notice that:\n\n   a b c a b (c+1) = (abc) * 1001 + 1\n\nSo the number needs to be a square which is 1 more than a multiple of 1001.\n\nIn number theory, we write this as:\n\n    z^2 == 1 (mod 1001)\n\nNow, we factor 1001 into primes as 1001 = 13 * 11 * 7\nand we can say that the above equation is the same as the three equations:\n\n    z^2 == 1 (mod 7)\n    z^2 == 1 (mod 11)\n    z^2 == 1 (mod 13)\n\nNow, for prime numbers, z^2 == 1 (mod p) can be solved only for z = +/- 1 (mod p)\n\nSo we get eight possible triples of equations:\n\n    z == (1,1,1)    (mod 7,11,13)\n    z == (1,1,-1)   (mod 7,11,13)\n    z == (1,-1,1)   (mod 7,11,13)\n    z == (-1,1,1)   (mod 7,11,13)\n    z == (1,-1,-1)  (mod 7,11,13)\n    z == (-1,1,-1)  (mod 7,11,13)\n    z == (-1,-1,1)  (mod 7,11,13)\n    z == (-1,-1,-1) (mod 7,11,13)\n\nNow, the equation:\n\n    z == (p,q,r)   (mod 7,11,13)\n\nis solved via the Chinese remainder theorem, and you get the following values for z:\n\n     z = 1,     z^2 = 000001\n     z = 155,   z^2 = 024025\n     z = 274,   z^2 = 075076\n     z = 428,   z^2 = 183184\n     z = 573,   z^2 = 328329\n     z = 727,   z^2 = 528529\n     z = 846,   z^2 = 715716\n     z = 1000,  z^2 = 1000000 (obviously, not a solution)\n\n[ You can think of the 1,000,000 case as when a=b=c=9. ]
562792
A=[1 2 3;3 4 5;7,8 9]\n\nA =\n\n     1     2     3\n     3     4     5\n     7     8     9\n\n\n>> B=(A([1 3],:))\n\nB =\n\n     1     2     3\n     7     8     9\n\n\n>> B=(A(:,[1 3]))\n\nB =\n\n     1     3\n     3     5\n     7     9\n\n\ngood luck i wish i could helped u
556722
look at my question\n\nBidDogARRRRR...\n1 minute ago\nDoes anyone know 0=1 and anything = anything else?\na = 1; b = 0\n1. a = b + 1 \n2. (a-b)a = (a-b)(b+1)\n3. a2 - ab = ab + a - b2 - b \n4. a2 - ab -a = ab + a -a - b2 - b\n5. a(a - b - 1) = b(a - b - 1) \n6. a = b\n7. b + 1 = b \n8. Therefore, 1 = 0
379711
Easy Homework?\n\na=2, b=1, k=2, j=5, n=3.\n2^2=1^2 (mod 3), and 2=5 (mod 3) but 2^5=2 (mod 3) and 1^5=1 (mod 3), so 2^5 is not 1^5 (mod 3).
939960
If P is an odd prime, then half of the time we have P = 1 mod 4, and the other half of the time we have P = 3 mod 4.\n\nThis is the same as saying that half of the odd primes are of the form 4n + 1, and the other half are of the form 4n + 3.
139883
(a+b)(a-b) \n= a^2 - ab + ba - b^2\n= a^2 - b^2 -ab +ab\n= a^2 - b^2
527089
Say a and b\na-b = 10 and a*b = 999\na = b+10 substituting this in another equation\n(b+10)*b = 999\nb^2+10b = 999\nb^2 + 10b - 999 = 0\nb^2 + 37b - 27b - 999 = 0\nb(b+37)-27(b+37) = 0\n(b+37)(b-27) = 0\nSo,\nb = -37 or b = 27\nand as a = b + 10\na = -27 or a = 37
187387
(n)(3n-1)(3n-2).......(n+1)
775741
I cannot prove it formally but I can mention the steps\n\na | u(n)\n\na | u(n+1)\n\nso a | u(n+1) - u(n) if a devided b and a devided c then it devides b-c\n\nso a | u(n-1)\n\nthere by we reach a | u(0) and a|u(1)\n\nu(0) does not give any infomation as u(0) =0\n\nso a | u(1) = 1\nso a= +/-1 \nu(0) = 0\nu(1)  = 1\n\nnext terms are u(2) = 1,u(3) = 2 ,u(4) = 3, then 5,8,13,
842954
the order of any element, modulo p, divides p-1 (which is an even number for everything except the stupid case of p=2, which i'm omitting) and every number relatively prime to p, raised to the (p-1) power will be congruent to 1 (modulo p)\n\ntherefore b^(p-1) = 1 (mod p)\nso a^((p-1)/2) = (b^2)^((p-1)/2) = b^(p-1) = 1 (mod p)\ntherefore order(a) divides ((p-1)/2) and hence order(a) does not equal (p-1), which is the requirement for being a primitive root.\n\nhence, a can never be a primitive root