Is there another pythagorean triple (x, y, z) except for (3, 4, 5) and triples like (3w, 4w, 5w)?
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Yes. Many. Here's a way to get more:\n\nPick and two numbers A and B with A more than B.\n\nLet x=A^2 -B^2\nLet y=2*A*B\nLet z=A^2 +B^2.\n\nThen it will turn out that x^2+y^2=z^2.\n\nAll primitive Pythagorean triples are of this form, by the way. Primitive means that there are no common divisors between the three numbers.\n\nFor example, if A=5 and B=2, then you get x=21, y=20 and z=29.
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Is this the Pythagorean theorem? There are many proper ways of proving it. Check wikipaidia below.
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Pythagorean triples are integer length sides of right triangles.\n\nIts in the form of (t^2 - s^2, 2st, t^2 + s^2)\n\nusing that can create another right triangle.\n\nSorry, but thats about all i could dig up. I'm not really all that familiar with P.T.
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The squares of two of the numbers must add to the square of the third.\nLets try x^2 = 24^2 + 51^2, then x^2 = 3177. Thats not a square.\nLets try 51^2 = 24^2 + x^2. Then x^2 = 2025, so x = 45.\nSo (24,45,51) is a pythagorean triple.
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Martin Gardner, for many years the recreational maths columnist in Scientific American magazine,\n\nHe was a great populariser. A puzzle he set interested me so much that it prompted me to discover the trinary tree of Pythagorean Triples, some ten years ago, in order to solve the problem.\n\nThe problem was, there is a right-angled field with one side 48, there are ten different integer solutions, what are they?\n\nThe first 8 are fairly easy being multiples of \n\n(3, 4, 5) (yields two solutions: 12 x the 4 and 16 times the 3)  \n(5, 12, 13) x 4\n(7, 24, 25) x 2 \n(8, 15, 17) x 6\n(12, 35, 37) x 4\n(16, 63, 65) x 3\n(24, 143, 145) x 2\n\nThe ninth solution is further down that series and is a primitive triple\n(48, 575, 577)\n\nBut the tenth eluded me. I therefore started to research the topic. I reasoned there might be a clue in the intervals between the 3 numbers in a triple. I took the triple (5, 12, 13) and wrote it as\n\n(-12) ^2 + (-5) ^2 = 13^2 and wondered if there was another triple with intervals 7 18 and 25 that wouid be three positive numbers if I solved the quadratic equation created by putting x, x+7, and x+25 into Pythagoras' Theorem:\n\nx^2 + x^2 + 14x + 49 = x^2 + 50x + 625\nSimplifying\nx^2 - 36x - 576 = 0\n(x - 48) (x + 12) = 0\n\nSo one triple is (48, 55, 73) and the other is (-12, -5, 13) which is what we started from.\n\nThe triple (48, 55, 73) gave me the tenth solution to Gardner's problem but also a method of generating all possible primitive triples from (3, 4, 5).\n\nBy writing 5, 12, 13 as\n(-5)^2 + 12^2 = 13^2, a similar search for a triple with intervals 17, 1, 18 leads to seeking a triple (x, x+17, x+18) and the quadratic equation\n\nx^2 + x^2 + 34x + 289 = x^2 + 36x + 324\nx^2 - 2x - 35 = 0\n(x -7) (x + 5) = 0\nwith triples (7, 24, 25) and (-5, 12, 13) with which we started\n\nSimilarly \nBy writing 5, 12, 13 as\n(-12)^2 + 5^2 = 13^2, a similar search for a triple with intervals 17, 8, 25 leads to seeking a triple (x, x+17, x+25) and the quadratic equation\n\nx^2 + x^2 + 34x + 289 = x^2 + 50x + 625\nx^2 - 16x - 336 = 0\n(x - 28) (x + 12) = 0\nwith triples (28, 45, 53) and (-12, 5, 13) with which we \n\nThese three techniques generate three new triples, from any triple and can be expressed as matrices e.g. \n\n(2 1 2\n 1 2 2    X (5, 12, 13) = (48, 55, 73)\n 2 2 3)
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3 to the 5th power\nor 3*3*3*3*3
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deep sea is right, but considering the fact we're talking basic arithmatic, not sure what your looking for, are you looking for something more advanced....  again answers are:\n\n1)3+3=6\n2)7*3=21\n3)51-48=3\n4)5*2+1=10+1=11\n5)10+5=15\n6)3*2=6\n7)4+2=6\n\nAgain, unless your 5 years old, I'm sure you already knew those...  I'm not sure what you mean by algebra.  Algebra is the substitution of a letter for an unknown quantity such as 2+x=4, where x=2, meaning 2+2=4\n\nAlso, this is your homework, not mine...  Remember nobody on YA is here to do your homework for you, althouogh I'll always be willing to give a few hints if needed...  Also people might give you wrong answers to mess you up...  Algebra isn't hard once you learn it, so have fun
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The "3, 4, 5" triangle is the most famous of the Pythagorean triples. It is the smallest right triangle with a right angle and integral side lengths.\n\nThe other four are: 5, 12, 13; 7, 24, 25; 8, 15, 17; and 20, 21, 29.
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(4,3) and (-4,-3)\nI think.\n\nCan't blame ya for asking about that one.  Trying to do that equation without a calculator before Calc III can be a pain since you don't get a shortcut in differentiating that type of  problem until late into Calc III.  It can be a pain even with a calculator without that shortcut due to the fact is doesn't balance easily either into a y=f(x) form.
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First, you should know that you can multiply any one of these equations through by a nonzero constant, without changing the solution. You can also add or subtract any two of the equations term by term without changing the solution.\n\nWe begin by adding equations 1 and 3 together\nx+y+z = -5\nx-y+4z=12\n---------------\n2x + 5z = 7  (notice that the ys cancel out)\n\nNow we triple equation 1, and subtract equation 2 from it\n\n3x+3y+3z= -15\n2x+3y-2z = 8\n--------------------\nx +5z = -23\n\nNow we have two equations, in which we have eliminated y.\nWe subtract them, and solve for x.\n\n2x + 5z = 7 \nx    +5z = -23\n-----------------\nx = 7 - (-23) = 30\n\nIt is easy to find y and z, now that we know x is 30/\n\nx + 5z = -23\n30 + 5z = -23\n5z = -53\nz = -(53/5) or -10.6\n\nx + y + z = -5\n30 + y + (-53/5) = -5\ny = 53/5 - 30 - 5 = -24.4\n\nSo (x,y,z) = (30,-24.4,-10.6)
9842
These numbers seem to be defined as : x,x+2,x+3 for every triplet. \nTherefore, 1,3,4,5,7,8,9,11,12,13,15,16,...
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uh..... no. Unless I am missing something, you substitute the 3 for 'x' and the 2 for 'y' and get 3+(3 x 2) = 3 + 6 = 9
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The Pythagorean Theorem? It's a^2 + b^2 = c^2.\n\nIf there's a right triangle, and a and b are the two "legs" (the shorter sides), and c is the hypotenuse (the longest side).
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The most intuitive way to answer is to use Gauss-Jordan elimination, as described below.\n\n\nw + x + y + z = 24\n3w + 3x + 3y + 3z = 72\n3w - 2x - 3y + 3z  = -9\n5x + 6y= 81\n\nw + x + y + z = 24\n5w + 5x + 5y + 5z = 120\n5w + 3x - 2y - 2z  = 4\n2x + 7y + 7z = 116\n\nw + x + y + z = 24\n2w + 2x + 2y + 2z = 48\n2w + 4x + 4y - 5z  = 22\n-2x - 2y + 7z = 26\n\n2x + 7y + 7z = 116\n-2x - 2y + 7z = 26\n\n4x + 9y = 90\n5x + 6y= 81\n\n20x + 45y = 450\n20x + 24y = 324\n21y = 126\n\ny = 6\n\n4x + 9y = 90\n4x + 54= 90\n4x  = 36\nx = 9\n\n2x + 7y + 7z = 116\n18 + 42 + 7z = 116\n7z = 56\nz = 8\n\nw + x + y + z = 24\nw + 9 + 6 + 8 = 24\nw = 1\n{w,x,y,z} = {1,9,6,8}\n\nAnother method is to put the coefficiants into a matrix, and put it into reduced-row echelon form.\n\n|1 1   1  1  24|\n|3  -2 -3 3 -9|\n|5  3  -2 -2 4|\n|2  4  4  -5 22|\n\nIn RREF...\n|1 0 0 0 1|\n|0 1 0 0 9|\n|0 0 1 0 6|\n|0 0 0 1 8|\n\nThese are all new valid equations, just in a much nicer form. Specially, 1w+0x+0y+0z=1, etc.\nThis tells us that w=1, x=9, y=6, and z=8.\n\nAlso, you could use Cramer's rule with these equations and get the same answer, but I will not demonstrate that one.
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If rhe two legs and the hypoteneuse are all integers, then the thgree numbers are known as a Pythagorean Triple. There are many properties of triples that help you to work out what the hypoteneuse is.
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y^2 + 7y + 3y + 21\ny^2 + (7 + 3)y + 21\ny^2 + 10y + 21\n(y + 7)*(y + 3)
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There are lots of them.  The story of Fibonacci numbers is interesting.  Also, some background on the Pythagoreans:\n\nhttp://en.wikipedia.org/wiki/Portal:Mathematics