This is a number theory question about congruences and primitive roots.
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Let G be the group of integers in { 1, ..., m } that are relatively prime to m, with the operation of multiplication modulo m.\n\nThen the subgroup < b > generated by b has h elements and the subgroup < c > generated by c has k elements.  Now the intersection of < b > and < c > is a subgroup of both < b > and of < c > (and of G, of course) so its size n divides evenly into both h and k.  But gcd(h, k) = 1 so < b > intersect < c > is the one element subgroup { 1 } of G.\n\nNow (bc)^(hk) = b^(hk) c^(hk) = b^(hk) c^(kh)\n = (b^h)^k (c^k)^h = 1^k 1^h = 1 in G (i.e., modulo m) so the order of bc modulo m divides evenly into hk.\n\nLet r = order of bc modulo m.\n\nSo (bc)^r = 1 = b^r c^r implies b^(-r) = c^r but the left hand side is in < b > and the right hand side is in < c > so since the intersection of < b > and < c > contains only the multiplicative identity it follows that b^(- r) = c^r = 1 and also b^r = 1.\n\nSo h divides r and k divides r.  Therefore lcm(h, k) divides r.  But hk = lcm(h, k) gcd(h, k) = lcm(h, k) 1 = lcm(h, k) so hk divides r.  But r <= hk, r = hk = order of bc modulo m.
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It is not known to be a fact.  However, it would be a special case of the Fermat-Catalan conjecture about integer solutions of x^p + y^q = z^r, and it has been proved that a similar special case to yours has no solutions.
671515
It's probably a mistyping of FLT which is shorthand for Fermat's Last Theorem. \n\nFLT states that threre are no integer values a, b, c that are solutions to the equation a^n + b^n = c^n when n > 2.
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It's not square *roots*... it is the series of PERFECT SQUARES!!! How many times are people going to ask this? And isn't it obvious? This is part of some Google DaVinci Code contest, isn't it?\n\nWhy don't you just search for the same question because it has been asked about 31 times now!!!
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1) (I had had another proof but I like this one better)\nlet x = 4k+a for k = 0,1,2,3,... and a = 0,1,2, or 3\nS = 1^x + 2^x + 3^x + 4^x = 1 + 2^(4k+a) + 3^(4k+a) + 4^(4k+a)\n= 1 + 2^a(2^4)^k + 3^a(3^4)^k + 4^a(4^4)^k \n≡ 1 + 2^a(1)^k + 3^a(1)^k + 4^a(1)^k (mod 5) by Fermat's theorem\n≡ 1 + 2^a + 3^a + 4^a (mod 5)\n= 10, 30, 100 ≡ 0 (mod 5) i.e., S divisible by 5 when a = 1,2,3 \n≡ 1+1+1+1 ≡  4 (mod 5) i.e., S not divisible by 5 when a=0, i.e, x divisible by 4\n \n2)\nx^2 - y^2 = 12^2 = 144\n(x+y)(x-y) = 144\nThe 15 factors of 144 are 1,144; 2,72; 3,48; 4,36; 6,24; 8,18; 9,16; 12\nx+y and x-y will be integers only when cofactors are both even or both odd.\nThis leads to (37+35)(37-35); (20+16)(20-16); (15+9)(15-9);\n(13+5)(13-5)...[i.e.,4 cases (x,y) = (37,35), (20,16), (15,9), (13,5)]\n\n3)\nIf x is even (x=2k), x^4 + 4^x = (2k)^4 + 2^(2x) and is divisible by 2, so not prime.\nIf x is odd (x=2k+1), then\nx^4 ≡ 1 (mod5) for any x by Fermat's theorem,\nand 4^(2k+1) ≡ 4*4^2k ≡ 4*2^4k ≡ 4*(2^4)^k ≡ 4*1^k ≡ 4 (mod 5)\nx^4 + 4^x ≡ 1+4 ≡ 5 ≡ 0 (mod 5) and is thus divisible by 5 , so not prime.\n\nI hope you understand these and are not just handing it in.\nIf there's a next time please ask three questions.
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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. ]
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So call the number\nB = {b_0,b_1,b_2,b_3,b_4,b_5,b_6,b_7,b_8,b_9}\n(this reverse indexing makes things easier).\n\nYou must have b_0 != 0, since you were told it was a ten-digit number. So b_0 is between 1 and 9.\nSo b_0 = 0 - (number of the b_i which are non-zero)\n\nAlso note that Sum_i=0..9 (b_i) = 10 since B has ten digits, and each b_i has to have a single value between 0..9\nHence Sum_i=1..9 (b_i) = (10 - b_0) for any particular choice of b_0\n\nTo prove the number is unique, see below:\n\nIt's easy to show 6210001000 is one possible answer.\nb_0 = 6 since there are six 0's in B\nb_1 = 2 since b_2, b_6 =1\nb_6 = 1 since b_0 = 6\nAll other b_i are 0 since there are no other non-zero b_i.\n\nTo prove the number is unique, the easiest way seems to be to show that the other b_0 between 1..5 and 7..9 are impossible by (proof by contradiction).\n\nb_0 = 9 is impossible since then b_9 must be 1, which gives a contradiction giving '9000000001', which only has 8 zeros, also b_1 must be 1, et.c\n\nb_0 = 8 gives '8000000010', but then b_1 must be 1, and it has one less zero, so that becomes '7100000100'...\nbut that has two ones, so b_1++, and b_2 must be 1, and you have again one less zero so b_0=6, giving '6210001000', which finally works.\n\nHarder to show that b_0 = 1..5 are impossible.\n\nLet's start from the opposite extreme, b_0=1.\nThen Sum_i=1..9 (b_i) = (10 - b_0) = 9\nThis constraint means each b_i can only be 0, 1 or 2.\nSince we set b_0=1, this means we have one zero and eight ones e.g. B='11111111110', doesn't matter where you put the 0. But that doesn't work since clearly b_1 must >=8, thus Sum_i=1..9 (b_i) = 8+7>9 which is a contradiction.\n\nIt should be trivial to crank out the cases b_0=2..4, maybe there is a more compact proof.
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It's a little tricky, but here goes:\n\nOk, assume that rt2 can be written as a rational fraction, then there must be some integer, "a", such that a*rt2 is an integer.  For example, if rt2 were 3/4, then "a" would be 4.  4*3/4 = 3, an integer.\n\nAlso assume that, whatever number "a" is, it is the smallest possible integer.  As in the example, 8,12,16 etc would also yield integers, but 4 is the smallest one possible.\n\nSo to recap, we have a*rt2 is some integer.\n\nNow we also know that 1<rt2<2.  Subtract 1 all the way across 0<rt2 - 1<1\n\nso now we know that rt2-1<1, and, logically, a*(rt2 - 1) < a*rt2\n\nTake a moment, let all of that sink in...and moving on:\n\na*(rt2-1) = a*rt2 - a  --- THIS WILL ALSO BE AN INTEGER, we'll assign it a new variable, b, and remember that above we proved that b is less than a.\n\nIf b = a*rt2 - a \n\nThen by multiplying both sides by rt2 you get\n\nb*rt2 = 2a - a*rt2\n\nNow, if a is an integer, and a*rt2 is an integer, then 2a-a*rt2 must also be an integer.  And since it is equal to b*rt2, then b must also be an integer.\n\nBut wait - we said that in a*rt2, a was the smallest possible integer, and we know that b is smaller than a, so there is your contradiction.\n\n\nPersonally, I like the other proof better I think.  This one is quite nice, but it was rather difficult to explain.  Once it sinks in, though, it is actually simpler.  I hope this helps.
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1. It's one multiplied by ten to the twenty-third power.\n  -Otherwise ten to the twenty-third power.\n\n2. It is also an example of multiplicative identity.\n\nSorry if that is not what you are looking for.
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To show that if a natural number is divisible by three, then the sum of its digits is divisible by three, use inductive reasoning:\n\n(A)  Base case: 3 is divisible by 3, as any natural number is divisible by itself.  The sum of its digits is 3, which is, of course, divisible by 3.\n\n(B)  Assumption: Let k=d(n)d(n-1)...d(1) be a multiple of 3, where the d(i) is the ith digit from the right in k.  Assume that d(n)+d(n-1)+...+d(1) is divisible by 3.\n\n(C)  Induction: Now, consider k+3.  One of the following occurs:\n\n     (1)  3 is added to d(1) and no carries are involved.  The sum of the digits becomes d(n)+d(n-1)+...+d(1)+3, which is divisible by 3.\n\n     (2)  3 is added to d(1) and a carry to the tens place is involved.  After the carry, 10 is subtracted from d(1) and 1 is added to d(2).  Now, The sum of the digits becomes:\n\n  d(n)+d(n-1)+...+d(1)+3-10+1\n= d(n)+d(n-1)+...+d(1)-6,\n\nwhich is divisible by 3.\n\n     (3)  3 is added to d(1) and multiple carries are involved.  This occurs when a string of m-1 9s ending in the tens place exists.  After the carries, 10 is subtracted from d(1) and 1 is added to d(2) (making d(2) 10), 10 is subtracted from d(2) (making d(2) zero) and 1 is added to d(3), and so on, involving m carries.  Then, Now, The sum of the digits becomes\n\n  d(n)+d(n-1)+...+d(1)+3-10m+m\n= d(n)+d(n-1)+...+d(1)+9m-6,\n\n which is divisible by 3.\n\nHence, by induction, if a natural number is divisible by three, then the sum of its digits is divisible by three.  \n\nNow, you wanted the converse.  Similar proofs can show that for base cases 1 and 2, the sum of digits for k=3*p+1 and k=3*p+2 (where p is a natural number) will not be divisible by three, giving the proof that a natural number is divisible by three if and only if the sum of its digits is divisible by three.  QED
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Proving irreducibility over Q is rather straightforward based on what the form of a nontrivial factorization would have to look like.\n\nReduction modulo every prime follows from X^4 - 10x^2 + 1 being equal to all three of the following:\n\nx^4 - 2x^2 + 1 - 8x^2 = (x^2 - 1)^2 - 2(2x)^2\nx^4 + 2x^2 + 1 - 12x^2 = (x^2 + 1)^2 - 3(2x)^2\nx^4 - 10x^2 + 25 - 24 = (x^2 - 5)^2 - 6(2^2)\n\nIf 2 is a quadratic residue modulo p, then the first is a difference of two squares.\n\nIf 3 is a quadratic residue modulo p, then the second is a difference of two squares.\n\nIf neither 2 nor 3 is a quadratic residue modulo p, then their product 6 is a quadratic residue modulo p, and so the third is a difference of two squares.
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A pythagorean primitive is a triple in which their only common divisor is 1.\n\nFor example: 3,4,5 is a Pythagorean primitive and 6,8,10 is not.
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b)\n\nNote that 314159266 is of the form 4k + 2, where k is an integer.  Since any integer is of the form 2n or 2n +1, this means that all squares of integers are either of the form 4m or 4m + 1.  No matter how we try to do the subtraction of two numbers of the form 4m or 4m + 1, we won't get a number of the form 4k + 2:\n\n4m - 4n = 4k\n4m - (4n + 1) = 4k + 3\n(4m + 1) - 4n = 4k + 1\n(4m + 1) - (4n + 1) = 4k.\n\nCheers!
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Ok, I think that I got it. on the third one you posted, it's about simplifying the binomials (2 numbers). So you add the exponents of the like variables (or coefficiants- there's no number, then it's assumed to be 1. Ex.\n(z^4)^-3 * (z^-2)^-5\n(z^2)^-8\n\nAre there no answers in the back of the book? I would email you, but it blocks your address, if I could see some of your notes it would be easier to explain I think.
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Let's say at the outset that non-primitive triples (eg (6, 8, 10) and (9, 12, 15) are not particularly interesting. What are interesting are primitive ones eg (3, 4, 5) in which the 3 members of the triple have no common factors. After all, any fool can multiply a primitive triple by 2, 3, 4 etc, what we are after when we go triple-hunting are primitive ones. \n\nAnd what we really want is a method of generating triples, so that gaving found some, we can make more. Preferably a method that only generates primitive triples, so we don't have to exercise quality control and weed out non-primitive ones afterwards.\n\nAnd like the Spice Girls, what we really, really want is a method that generates triples from triples not from generating numbers that don't have any aparent connection with triples being plugged into a formula. Preferably a method that can generate all possible primitive triples just from (3, 4, 5), knowing only that one and no other. That would be so much more elegant and satisfying, especially if some findamental structure and relationships between triples were thereby revealed.\n\nThere are an infinite number of triples. So we want a method that generates the lot, not just a subset of them.\n\nAnd triples have many unusual and unexpected properties, making them rewarding to study, if you like finding patterns in nature. This can create a sense of awe and wonder and beauty, in addition to the aesthetic pleasure of elegance of solution.\n\n====\n\nAt the age of 5 I discovered two simple rules for generating primitive triples. As I recall, I only knew (3, 4, 5) and wondered if there were any more like that. Curiosity got me started. I knew no algebra at five.\n\n(1) take any odd number, square it, halve that, take the numbers either side.\n\nThus 3 => 9 => (3, 4, 5)\n5 => 25 => (5, 12, 13)\n7 => 49 => (7, 24, 25)\n9 => 81 => (9, 40, 41) \n11 => 121 => (11, 60, 61) etc\n\nAll such triples have their two largest sides differ by 1. Note how the other terms advance in an Arithmetic Progression, in steps of 8, 12, 16, 20 etc.\n\n(2) Take any even number divisible by 4, square it, divide that by 4 and take the numbers either side\n\nThus 4=> 16 => 4 => (4. 3. 5)\n8=> 64 => 16 => (8, 15, 17)\n12=> 144 => 36 => (12, 35, 37)\n16 => 256 => 64 => (16, 63. 65)\n20 => 400 => 100 => (20, 99, 101) etc\n\nNB Even numbers that are NOT divisible by 4 (6, 10, 14 etc) will generate triples but they are not primitive triples. And it doesn't work for 2.\n\nAll such primitive triples have their two largest sides differ by 2. Note how the other terms advance in an Arithmetic Progression, in steps of 12, 20, 28, 36 etc.\n\nHaving worked out these two rules for generating triples, in my naivete, I thought that was all there was to it. That these were the only triples in existence.\n\nI told parents and teachers, who didnt know of any other triples and weren't able to help me develop my theories.\n\nNobody told me about (20, 21, 29). If they had have done I would have realised that whilst there was a difference of 1 between two sides here, it was between the two shortest sides, not the two longest sides, so there was almost certainly a series of such triples to be discovered. and more than was dreamed of in my philosophy.\n\nNote that 20 x 20 = 400, divide by 16 = 25 and then take two numbers that are 8 apart either side => (20, 21, 29) so it has similar properties. \n\nIndeed there is such a series, the next triple in it is (119, 120, 169). \n\nNote that 120 x 120 = 14400, Divide by 100, and get 144, take two numbers that are 50 apart, either side and voila we get 119, and 169, so similar properties exist here, too.\n\nYou can generate each from the previous one in the series, and each new triple has legs that are about 6 times the size of the previous one. And the hypoteneise is about 1.41 times the lengths of the legs as these values converge o
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We assume that Sum and Prod know what we know.\n\nFrom the rules, xy is not prime.\nFrom Prod's first statement, x and y are not both prime (although one of the two could be)\nFrom Sum's first statement we know that x+y is a sum none of whose possible components are both prime (x+y is not even), but from her last statement, it must have more than one possibility - this rules out any x+y > 36.\nThis shaves x+y down to 11,17,23,29, or 35.\n\nChecking 11, we see that if x=2 and y=9, Prod wouldn't be able to initially deduce the numbers from knowing xy = 18, thinking they could be 2,9 or 3,6. This solution runs into a problem when we realize that Sum could still be thinking that x,y could be 3,8.\n\nNow we have to exclude all the other possibilities. All the other numbers possible for x+y have multiple solutions except for x+y = 35\n\nAt first Sum is thinking 2*33=66 or 3*32=96. Either way Prod can't deduce on her own. Once Prod hears Sum's statement Prod is able to narrow it down. Prod could not have 66, or Prod would still think the Sum could be 2+33=35 or 6+11=17. Therefore Prod must have 96 allowing them both to deduce the numbers 3 and 32.
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A:\n\nA hint for these is that a root of a number raised to the power is the number raised to (power/root)\n\ne.g. cube root of x^4 = x^(4/3)\n\nB(i):\n\nEasy - just replace x with 1, 2, 3, 4 and 5 and work out the answer\n\nB(ii):\n\nSimilar to B(i)