Why is x^2+y not reducible?
413310
Irreduicible basically means that you can't factor it.  This particular one can't be expressed as the product of two or more irreduicible polynomials.
814311
The issue lies in the fact that the original equation is 0=0.\n\nThen, if you put in any value for x, you are dividing by 0 which as we all know, is a "no no."
471125
Hi!\nI'm suprised nobody used the easiest way that i will demonstrate now.\n\nLets assume further that x is positive real.\n\nWe start with a simple fact that ANY squares are nonnegative. The idea is we "choose" a square intelligently to suit our purpose. Also not that x does NOT equal zero.\n\nBy experience, \nI choose the square ( x^(1/2) - 1 / [ x^(1/2)] )^2.\nIt must be nonnegative for whatever positive real values of x.\nThus,\n\n    ( x^(1/2) - 1 / [ x^(1/2)] )^2 >= 0\n\nExpanding and regrouping, we get,\n\n    x - 2 + 1/x >= 0 \n      \nor equivalently,\n\n      x + 1/x >= 2\n\nThus, minimum is 2. \nWith equality when x^(1/2) = 1 / [ x^(1/2)], or\n x = 1.\n\nDone.\n\nAlternative Method.\n\nOne can use two variable AM - GM,\nFor any two positive reals m and n, we have\n\n ( m + n )/ 2 >= (mn)^(1/2)\n\nWith equality if and only if m = n.\n\nJust set m = x and n = 1/x, to get,\n\n( x + 1/x )/ 2 >= 1\n\nor  x + 1/x >= 2\n\nAgain, the minimum values is 2 with equality if and only if x = 1 .\n\nRegards
67585
Formulaically it reduces down to x*z = x*z, so no matter which numbers you plug in for any of the values it will be a valid equation.  \n\n(x*(y+z))-(x*y)=x*z      becomes\nx*y + x*z - x*y = x*z     rewrite that as \nx*y + x*z + -(x*y) = x*z    the positive and negative x*y terms cancel,leaving\n\nx*z = x*z\n\nSince that is a truism, an infinite number of solutions exist.
413711
Assume x^2 + y^2 = 8z + 3 for some positive integers x, y, and z.\n\nx^2 + y^2 = (x+y)^2 -2xy = 8z + 3\n\n(x+y)^2 -2(xy+4z) = + 3\n\nCompleting the square gives:\n(x+y)^2 -2(xy+4z) + (xy+4z)^2 = (xy+4z)^2 + 3\n((x+y) - (xy+4z))^2 = (xy+4z)^2 + 3\n\nHere you have two squares of positive integers whose difference is 3.  These two squares must be 4 and 1.\n\nHence, (xy+4z) = 1 and (x+y) - (xy+4z) = 2, which gives (x+y) = 3.  \n\nGiven that x and y are positive intergers, this implies that\n\n1^2 + 2^2 = 8z + 3\n5 = 8z + 3\nz = 1/4, which is not an integer, contradicting our previous assumptions.
1172077
Your claim that there are infinitely many solutions for x and y for a given x and n does not hold if we consider only integer solutions.\n\nConsider that we have:\nx^n = z^2-y^2\nx^n = (z-y)(z+y)\nSince z, y are integers, |z-y| >= 1 and |z+y| >= 1\nAlso, |z-y| <= x^n and |z+y| <= x^n\nWe can assume that z,y > 0, since if there is a solution with one of them less than 0, then switching the sign will still be a solution.\nSo, now we know that |z+y| <= x^n.  For z,y >=0, this has x^n + 1 solutions (not all of these will solve the original equation, but there are no others that do).\n\nThus, there are at most 4x^n +4 unique solutions to this equation in the integers.  The others are found by taking a positive pair (z,y) and noticing that (-z,y) (z,-y), (-y,-z) are also solutions.\n\nHowever, if you are talking about non-integer solutions, then there are obviously infinitely many, and that result is not useful.
863763
D ( 2 or 0)\n\nThis is because the coefficients of the given cubic polynomial are real numbers and so any real zero needs be accompanied by its complex conjugate and so as stated  earlier MUST come in pairs so either there are 2 or there cannot be any.
402130
I get that the irreducible monics, with the primitive ones so marked, are:\n\nx^3 + 0 x^2 + 2 x + 1 (primitive)\nx^3 + 0 x^2 + 2 x + 2\nx^3 + 1 x^2 + 0 x + 2\nx^3 + 1 x^2 + 1 x + 2\nx^3 + 1 x^2 + 2 x + 1 (primitive)\nx^3 + 2 x^2 + 0 x + 1 (primitive)\nx^3 + 2 x^2 + 1 x + 1 (primitive)\nx^3 + 2 x^2 + 2 x + 2\n\nYour mileage may vary. :-)\n\nDegree three reducibles factor in one of the two forms:\n\n(x - a)(x - b)(x - c)\n(x - a)(x^2 + bx + c) with x^2 + bx + c irreducible\n\nSo if p is a degree three reducible then at least one of p(0), p(1), and p(2) must be 0 modulo 3.  Applying that test leaves the eight monic irreducibles shown above.\n\nThe primitive ones are found by verifying that x + <p(x)> has multiplicative order 26 in (Z/3Z)[x].
739976
Add all the polynomials together and you get:\nx^4 y^4
859279
This would be a special case of the Fermat-Catalan Conjecture.\n\nThis conjecture is that there are only a finite number of solutions of x^p + y^q = z^r for 1/p + 1/q + 1/r < 1.  Ten solutions are known, but all of them have the three exponents different (except for 1^p + 2^3 = 3^2) and one exponent equal to 2.\n\nYour conjecture would be the special case that there are no solutions with p > 2 and q = r.  It is already known that there are no solutions with p = q and r > 2 (Darmon and Merel, 1997).
697620
x^2+y^2=(x+y)^2-2xy\nsubstituting\nx^2+y^2=3^2-2(1)\n            =9-2\n            =7\nx^2+y^2=7\nmy 10 pointsplease
700473
(x^5 + y^5) =  (x+y)(x^4 - x^3*y + x^2*y^2 - x*y^3 + y^4).
1031270
x^2 + 5 + 5x + x y +5y\n\nx^2 + xy + 5x + 5y + 5\n\nx(x+y) + 5(x+y) + 5\n\nx(x+y) + 5(x+y+1)       OR     (x+5)(x+y)+5\n\nCynthia
858227
Multiply it out, mayhaps.\n\nx^2 +x + y - y^2\n\nWhich is also\nx(x + 1) + y(1-y)\n\nWhy does this question have four answers below, ALL OF WHICH STATE THE SAME THING?
908083
Tried to do it algebraically, that was a booger.\n\nI plugged that into mathematica and it gave me an error normally generated when the inverse of a function is needed to represent a solution and the solution variable occurs in more than one argument of that function.\n\nThat one might not have a solution?
665576
x^2<4x\n\ncase 1: x<=0\nno solution\n\ncase 2: x>0\ndivide by x\nx<4\n\ntherfore the solutions are 0<x<4
122740
you need to use the formula for (x+y)^2 and then use the numbers:\n(x+y)^2 = x^2 + 2xy + y^2 = x^2 + y^2 + 2*xy\nwe know that x^2 + y^2 = 7 and xy = 5\nso (x^2 + y^2) + 2*xy = 7 + 2*5 = 7 + 10 = 17\n(x + y)^2 = 17