Prove that there are no integers, x, y, z, such that x^2+y^2=8z+3.?
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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.
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numbers 3 and 4\n\nx(squared) is still an integer and xy/x=y, which is an integer
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Fermat's last theorem is that there is no solution in positive integers the the equation x^n + y^n = z^n, with n>2. \n\nThe above equation is different because some of the terms are raised to the 2nd power instead of the nth power. It is also different because it asserts an infinite number of solutions and Fermat's theorem asserts no solutions.
580657
Fermat's Last Theorem states that, for an integer n > 2, there are no non-zero integers x,y,z such that x^n + y^n = z^n.  What you wrote above does not prove this.  Yes, we know that 3^2 + 4^2 = 5^2, but there are no integers such that x^3 + y^3 = z^3.  If you still think your explanation actually proves this, please explain more.\n\n"There lies a commmon factor between a,b,c when a=b+c".  For your example, 9+16=25, the common factor is 1.  This does not, in any way, prove FLT.  You cannot just end a proof with a vague statement and not show how it proves what you're trying to prove.  How do the two facts 9+16=25 and "common factor when a=b+c" relate to prove FLT?\n\nUpdate:  I think you are confused as to what FLT is.  I stated it correctly above and the link below has a great explanation.  I believe you're trying to prove that there are an infinite number of integers x,y,z such that x^2 = y^2 + z^2.  Of course this is true, and has been known for a long time.  They're called Pythagorean triples.  FLT excludes the case n=2.  Read the link below for more information.
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Wow, what an interesting question. I'm surprised no one has solved it yet, but that may be a consequence of the fact that no one has understood it yet. Clearly, schools need to focus more on reading comprehension ^_^.\n\nTo answer your question - not only can we prove that there exist integers a and b satisfying your assumptions, but we can even show a method of finding them. The proof is long, so sit back, relax, grab your favorite drink, and prepare to read.\n\nFirst, consider that any solutions of these equations must obey the quadratic formula: x=(-b±√(b²-4ac))/2a. Substituting the coefficients of the  actual equations in, we find that the solutions of the first equation are (-p±√(p²-4q))/2 and the solutions of the second equation are (-p±√(p²+4q))/2. Let us consider only the greater of the solutions in each pair - that is, replace each ± with a +. The difference between these solutions is then (-p+√(p²-4q))/2 - (-p+√(p²+4q))/2. Call this number K. We can then perform some simple algebra as follows:\n\n(-p+√(p²-4q))/2 - (-p+√(p²+4q))/2=K\nmultiply both sides by 2 and simplify\n√(p²-4q)-√(p²+4q)=2K\nsquare both sides\n(p²-4q) - 2√((p²-4q)(p²+4q)) + (p²+4q)=4K²\nsimplify\n2p²-2√(p^4-16q²)=4K²\ndivide both sides by 2\np²-√(p^4-16q²)=2K²\nsubtract 2K² and add √(p^4-16q²) to both sides\np²-2K²=√(p^4-16q²)\nsquare both sides\np^4-4K²p²+4K^4=p^4-16q²\nmultiply both sides by -1 and simplify\n16q²=4K²p²-4K^4\ndivide both sides by 4\n4q²=K²p²-K^4\nadd K^4 to both sides\nK^4+4q²=K²p²\ndivide both sides by K² and rearrange\np²=K²+4q²/K²\n\nThus, if we let a=K and b=2q/K, then p²=a²+b² as required by the problem. All that remains is to show that a and b are nonzero integers.\n\nLemma 1: K is an integer\nProof: All solutions of both quadratic equations are integers. K is the difference of two of these solutions - that is, K is the difference of two integers. Therefore, K is an integer.\n\nLemma 2: K is nonzero\nProof: K is the difference between a solution of the first equation and a solution of the second equation. Suppose K=0. Then these equations have a common solution - that is, for some x, x²+px+q=0=x²+px-q. This would imply that q=-q. This can only be true if q=0. But q≠0. Therefore, the equations do not have a common solution and K≠0.\n\nLemma 3: 2q/K is an integer\nProof: Earlier we showed that √(p²-4q)-√(p²+4q)=2K. We can perform some simple algebra as follows:\n\n√(p²-4q)-√(p²+4q)=2K\nmultiply both sides by (√(p²-4q)+√(p²+4q)) and simplify\n2K(√(p²-4q)+√(p²+4q)) = (p²-4q)-(p²+4q)\nsimplify further and rearrange:\n-8q=2K(√(p²-4q)+√(p²+4q))\ndivide both sides by -4:\n2q=-K(√(p²-4q)+√(p²+4q))/2\ndivide both sides by K:\n2q/K=-(√(p²-4q)+√(p²+4q))/2\n\nNote that 2K=√(p²-4q)-√(p²+4q) Adding 2√(p²+4q) to both sides of that identity gives 2K+2√(p²+4q) = √(p²-4q)+√(p²+4q). We can then substitute to get:\n\n2q/K=-(2K+2√(p²+4q))/2\nSimplifying at last yields:\n2q/K=-K-√(p²+4q)\n\nWe already know K is an integer, so if √(p²+4q) is also an integer, then 2q/K is the difference of two integers and therefore an integer. Now we know that the solutions of the second equation are (-p+√(p²+4q))/2 and (-p-√(p²+4q))/2. The difference between them is √(p²+4q). However, since both of the solutions to the second equation are integers, their difference must be an integer. Therefore √(p²+4q) is an integer, and by extension, 2q/K is an integer. Q.E.D\n\nLemma 4: 2q/K is nonzero\nWe know 2q/k=-K-√(p²+4q). Suppose 2q/K=0. Then -K-√(p²+4q)=0 From this we would get:\n\n-K-√(p²+4q)=0\nadd K to both sides and rearrange:\nK=-√(p²+4q)\nmultiply both sides by 2\n2K=-2√(p²+4q)\nremembering 2K=√(p²-4q)-√(p²+4q), we substitute:\n√(p²-4q)-√(p²+4q)=-2√(p²+4q)\nadd √(p²+4q) to both sides and rearrange:\n-√(p²+4q)=√(p²-4q)\n\nHowever we specified in the beginning t
92952
(8x+3)(4x+3)\n32x2+24x+24x+9\n32x2+48x+9
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Let 4^x = z.\nThen the given equation becomes z - 3/z = 8\nor z^2 - 3 = 8z [multiplying both sides by 'z']\nor z^2 - 8z - 3 = 0\n\nUsing Sridharacharya's formula for root of quadritic equation\nz = [-(-8) +/-sqrt{(-8)^2 - 4*1*(-3)}]/(2*1)\n= 8.358898943 or -0.358898943\n\nHence either 4^x = 8.358898943 or 4^x = -0.358898943\n\nIn the first case taking logarithm of both sides w.r.t. base 10\nlog(4^x) = log(8.358898943)\nor x*log(4) = log(8.358898943)\nor x = log(8.358898943)/log(4) = 1.53165646\n\nIn the second case we find that power of a positive number (4) is negative (-0.358898943) which is impossible.\n\nSo 4^x = = -0.358898943 is inadmissible.\n\nHence x = 1.53165646 is the only solution.
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You know there are 8 axioms in mathematics which we cannot prove\n\n1) if 'a' and 'b' are integers then 'a+b' is an integer and 'a.b' is an integer (Closure law)\n2) a + b = b + a and a..b = b.a (cumulative law)\n3) a + (b + c) = (a + b) + c (Associative law)\n4) a (b + c) = a.b + a.c (Distributive law)\n5) a + 0 = a = 0 + a (Additive Identity)\n6) a.1 = a = 1.a (Multiplicitive Identity)\n7) for all integer 'a' there is a unique integer (-a) such that a + (-a) = 0\n8) I can't remember...\n\nSo we can prove your requirement using above axioms, \n\na + (-a) = 0\n(-1)(a + (-a) = -1(0)  (Multiplying by same factor)\n-a + -(-a) = 0 (Distributive law over addition)\n\nadding +a both sides\n\na + (-a + -(-a)) = a + 0 \n(a+ (-a)) + -(-a) = a (Associative law)\n0 + -(-a) = a \n-(-a) = a (proved... huh...)
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Concept:\nEvery even integer may be expressed in the form 2n, where n is an integer.\nEvery odd integer may be expressed in the form 2n + 1, where n is an integer.\n\nLetting x² and y² be the 2 perfect squares, their difference is:\nx² - y²\nIf we let y = x - 1, then\nx² - (x - 1)²\n= x² - (x² - 2x + 1)\n= x² - x² + 2x - 1\n= 2x - 1\n= 2x - (1)\n= 2x - (2 - 1)\n= 2x - 2 + 1\n= 2(x - 1) + 1\n\nSince x² is a perfect square, x is an integer and also x - 1 is an integer. Notice that we've converted it into the odd-integer form. Therefore, every odd integer is the difference of two perfect squares^_^
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1.y+2\n2. x+3\n3. z+6\n4. i wasnt sure what it was equal to, so i assued 0... making the answer (x+2)(x+5) however,if it is x2+7x=10...then this answer's wrong...
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Well, the basic problem you've run into is that the conjecture isn't true.  If a = 6, b = 3, and c = 9, then a*b = 18, and 18/9 = 2, but neither 6 nor 3 is divisible by 9.\n\nTo prove that something is true you must prove it is true for all possible values of the variables, in this case a, b, and c.  However, to prove something is false, you need only supply a single counter-example.  Thus the above example disproves the conjecture that for all integers a, b, and c if a*b is divisible by c then a or b is also divisible by c.\n\nNote that a ≠ cm and b ≠ cn does not imply that ab ≠ c(cmn).  For example, if a = 6, b = 8, c = 2, m = 4, and n = 3, then cm = 2*3 = 8 ≠6 so cm ≠ a, and cn = 2 * 3 = 6 ≠ 8 so cn ≠ b.  However, (cm)(cn) = 2 * 4 * 2 * 3 = 48, and a * b = 6 * 8 = 48, so ab = (cm)(cn).  This proves that you cannot multiply not-equalities (i.e. statements with "doesn't equal).\n\nAlso, concluding that if c does not divide a or b implies that c does not divide ab is not the same as concluding that if c does divide ab it must also divide a or b.  This logical fallacy is equivalent to the statement that for all plants, not a tree implies not a rock, so being a rock implies being a tree.  The first statement is true: all plants that are not trees are also not rocks.  However, the second statement is clearly false, as no rocks are also trees.
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No solutions my ascii, haven't these people ever heard of complex numbers?\n\nAs has been previously mentioned, for the first part of the equation you can set U=e^x and solve for U. Thus:\nU^2+6U+8=0\n(U+2)(U+4)=0\nU=-2 or -4\ne^x=-2 or -4\n\nNow, here's where we demonstrate that this equation has solutions. Remember Euler's identity: e^(iπ)+1=0. Thus, e^(iπ)=-1. Thus:\ne^x=-2\n(-1)e^x=2\ne^(iπ)e^x=2\ne^(x+iπ)=2\nx+iπ=ln 2\nx=ln 2+πi\n\nAnd from e^x=-4, we get x=ln 4+πi. So x=ln 2+πi or  x=ln 4+πi. And that's your solution.\n\nActually, for the sake of completeness, since e^(2πi)=1, x could also be any of these solutions plus an integral multiple of 2πi. Thus, x=ln 2 + (2K+1)πi or x=ln 4 + (2K+1)πi, where K is an arbitrary integer.
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You are solving for k so it you need to rewrite the problem to look like this k=z/(xy) where x=3, y=8, z=6 substitute the letters for numbers so k=6/(3*8) k=6/24 or k=1/4 to check just solve for any other letter such as z by using what you know of k and you should come out with z=6
271990
x^2-2xy+4xz-8yz=\n(x^2-2xy) + (4xz-8yz)=\nx(x-2y) +4z(x-2y)=\n(x-2y)(x+4z)
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Any method of solving will give x=0 y=0 and z=0 !
271727
x^2 + 12x + 27 = 0\n(x+9)(x+3)=0\nso...\nx+9=0\n  x=-9\nand\nx+3=0\n  x=-3
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2x+2y+3z=-1 .....(1)\n3x-1y-2z=18 .......(2)\n-3x+1y-1z=-9 ......(3)\n\n(1)+(2)x2\n2x+2y+3z=-1\n6x-2y-4z=36\n-------------------+\n8x-z=35 .............(4)\n\n(2)+(3)\n3x-1y-2z=18\n-3x+1y-1z=-9\n---------------------+\n-3z=9\nz=-3\n\nsubstitute this value of z to (4)\n8x-z=35\n8x-(-3)=35\n8x+3=35\n8x=32\nx=4\n\nsubstitute the value of x and z to (1)\n2x+2y+3z=-1\n2(4)+2y+3(-3)=-1\n8+2y-9=-1\n2y-1=-1\n2y=0\ny=0\n\n(x,y,z)=(4,0,-3)\n\nhope this is clear enough