1) Show that if p >=5 is a prime number, then p^2 + 2 is composite (hint: p takes one of the forms 6k+1 or 6k
336691
Wow this is a long question.... hehe.\n\n1) Firstly consider any integer divided by 6, we must get one of the residues ( remainders ), r = 0,1,2,3,4,5. If the remainder of y divided by 6 is r, thus we can write \n y = 6 x "quotient" + "remainder" = 6k + r, where let k denote the value of the quotient. Since r = 0,1,2,3,4,5, we have just shown that any integer y ( we may assume positive ), can be written in one of the "forms"\n y = 6k , 6k + 1, 6k + 2, 6k + 3, 6k + 4, 6k + 5.\n\nNow among these, the ones that are composite ( NOT prime ) \nfor any integer k, are\n      6k , Even\n    6k + 2, Even\n    6k + 3 = 3(2k + 1) , a multiple of 3. \n    6k + 4, Even\n\nRemembering that the only even prime number is 2.\nThus, the only possibilities for forms for primes in here is either 6k+1 or 6k + 5 . All other forms denotes a composite integer. Remember that a composite integer is an integer with at least two distinct prime factors.\n\nIf p is form 6k+1, Let p = 6k+1,\nThus, p^2 + 2 = ( 6k + 1 )^2 + 2\n              = 36k^2 + 12k + 3\n              = 3 ( 12k^2 + 4k + 1 )\n              = A multiple of 3 > 3 \n              = A Composite number.  \n\nIf p is form 6k+5, Let p = 6k+5,\nThus, p^2 + 2 = ( 6k + 5 )^2 + 2\n              = 36k^2 + 60k + 27\n              = 3 ( 12k^2 + 20k + 9 )\n              = A multiple of 3 > 3 \n              = A Composite number.\n\nThus, in both cases, if p is prime then p^2 + 2 is always composite.\n\n2)\nAssume that the number of primes are finite.\nNow the integer N is NOT divisible by any of the primes in the list p_1, p_2, ..., p_n.\nWhy? because say p_k divides N. Then, p_k must divide the following integer : p_1*...*p_n where it does NOT contain p_k. Which is a contradiction .\nThus, primes  must be infinite for if not, they produce the contradiction shown above.\n\n3)\n\nI will not use bertrand's conjecture because its NOT proven yet... So, i'll skip this one... its redundant. I will not  use something that has not been proven yet... i.e, a conjecture.
774052
LetP(n) denote the statement\n1^2+2^2+3^2+......+n^2=n(n+1)(2n+1)/6\nwhen n=1,P(1)=1(2)(3)/6=1 checks\nlet P(k) be true\ni.e.1^2+2^2+3^2+.......+k^2=k(k+1)(2k+1)/6\nto prove\nP(k+1)=(k+1)(k+2)(2k+3)/6 is true\nP(k+1)=1^2+2^2+3^2+.......+((k+1)^2\n=>[1^2+2^2+3^2+......+k^2]+(k+1)^2\nnow 1^2+2^2+3^2+......+k^2=k(k+1)(2k+1)/6\nsubsttituting\n=k(k+1)(2k+1)/6 +(k+1)^2\n=(k+1)[k(2k+1)+6(k+1)]/6\n=(k+1)(2k^2+7k+6)/6\n=(k+1)(k+2)(2k+3)/6\nhence proved\n\n2.try by your self
898910
If you mean a number with three prime factors, it's a 3-almost prime.  If you mean a number with three integral factors (prime, composite, or 1), it's the square of a prime.\n\nList of 3-almost primes: http://www.research.att.com/~njas/sequences/A014612\nList of squares of primes: http://www.research.att.com/~njas/sequences/A001248
830444
d.) 2m^2(6m^6+3m^4+7)
1178350
P=x^2+5x-50\nP=(x-5)*(x+10)\n\nif x=5 then \n\nP=(5-5)*(5+10)\nP=0*15\nP=0\n\nso x has to be greater than 5\nx = >5
175889
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.
643237
2 is the only (and hence also the highest) even prime number.\n\nOther even numbers would be divisible by 2 besides itself and 1, so they cannot be prime.
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.
332593
They are all either product of two prime numbers (35=5*7) or cubed prime (8=2^3).
890978
To get prime factors, keep dividing by numbers and divide those until you can't divide them no more!  For example,\n\n34 \n17 * 2\n\n(Can't divide 17 any more and can't divide 2 -- both of these are prime numbers)\n\nAnother example,\n\n120\n12 * 10\n2 * 6 * 2 * 5\n2 * 2 * 3 * 2 * 5 (<-- all prime numbers here)\n\nTry to get the rest yourself and post the answers, we'll tell you if you got them right!  ;-)
545916
4, 9, 25, 49\nPerfect squares such that the square root is a prime number.
749230
Neither.  A prime number has exactly two distinct factors, and 1 has only one.  And obviously it isn't composite.\n\nSome people will say 1 is prime, but they are wrong.
236680
A perfect square is the product of a whole number times itself, therefore, there is at least one whole number (its square root) which evenly factors into the number.  Since that is the case, the number can't be prime.\n\nOr you could argue the other way, a prime number is a number for which there is no other whole number (other than itself and 1) which can be a factor of it.  Since a perfect square has at least its square root as an factor, a prime cannot be a perfect square.
1165002
This is a wild guess.\n\n3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 = the number.\nThe answer will be the product of the first 12 primes that aren't 2. (If 2 were a factor, the answer would be even, not odd.)\n\nFor the rest, use a calculator. And check to make sure all those numbers are prime. Sorry this is such a whack answer.
806711
YES it is and here is a list of the firs 1000 prime numbers so just print it out and look at it while your homework.\nhttp://primes.utm.edu/lists/small/1000.txt
594239
ok so you have p^2 + p, you can factor out p from both parts of the expression, it will end up like this: p(p + 1).  Hope that works.
793441
[5-sqr(3)][8+sqr(7)] = 40 +5sqr(7) -8sqr(3) -sqr(3)sqr(7)\nSimplify:\n40 + 5sqr(7) - 8sqr(3) - sqr(21) and this should be your answer as 7 and 3 are prime