document.write( "Question 1197922: How many integer solutions (a,b,c) make the equation true? \n" );
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Algebra.Com's Answer #831391 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "If a,b,c are nonnegative integers where \n" ); document.write( "(a,b,c) = (0,0,17) \n" ); document.write( "(a,b,c) = (0,8,15) \n" ); document.write( "(a,b,c) = (1,12,12) \n" ); document.write( "(a,b,c) = (8,9,12) \n" ); document.write( "These four solutions are found through trial-and-error. Note that the second class (0,8,15) is from the pythagorean triple 8,15,17 which gives the equation 8^2+15^2 = 17^2 from the pythagorean theorem.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let's remove the requirement that \n" ); document.write( "So we can permute the elements, meaning something like (0,0,17) leads to (0,17,0) and (17,0,0). There are 3 such permutations.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let's now remove the requirement that the numbers a,b,c must be nonnegative. \n" ); document.write( "That yields twice as many solutions to give us 3*2 = 6 different solutions just based on the class (a,b,c) = (0,0,17). This is because the 17 could represent +17 or -17.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The class (0,8,15) has 3! = 3*2*1 = 6 permutations \n" ); document.write( "The 8 could be +8 or -8; the 15 could be +15 or -15 \n" ); document.write( "So we multiply by 2*2 = 4 to get 4*6 = 24 different solutions from the class (a,b,c) = (0,8,15)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The class (1,12,12) has 3! = 6 permutations if we could tell the '12's apart; however we cannot. So we divide by 2 to fix this erroneous double-counting to get 6/2 = 3 permutations instead. \n" ); document.write( "Each item could be positive or negative, so we multiply by 2^3 = 2*2*2 = 8 to get 8*3 = 24 different solutions of the class (a,b,c) = (1,12,12)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Lastly, the class (a,b,c) = (8,9,12) has 3! = 6 permutations. Each shows up 2 times to get us 2*2*2*6 = 8*6 = 48 different solutions here.
\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "From here we add up the frequencies to get \n" ); document.write( "6+24+24+48 = 102 \n" ); document.write( "which is the number of integer solutions to \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answer: 102 \n" ); document.write( " \n" ); document.write( " |