My Math Forum A really hard math problem "Prove that if..."

 Algebra Pre-Algebra and Basic Algebra Math Forum

 September 21st, 2013, 02:19 PM #1 Newbie   Joined: Sep 2013 Posts: 5 Thanks: 0 A really hard math problem "Prove that if..." Prove that if a b c are integers and satisfy the equation (a + 3)^2 + (b + 4)^2 ?(c + 5)^2 = a^2 + b^2 ? c^2 the common value of both sides is a perfect square. I would be very grateful if someone can help me
 September 21st, 2013, 04:01 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond Re: A really hard math problem "Prove that if..." Expanding the left hand side and canceling like terms gives 6a + 8b - 10c = 0, so a and b are multiples of 5. $c\,=\,\frac{3b\,+\,4a}{5}$ $a^2\,+\,b^2\,-\,c^2\,=\,a^2\,+\,b^2\,-\,$$\frac{3b\,+\,4a}{5}$$^2\,=\,a^2\,+\,b^2\,-\,\frac{9a^2\,+\,24ab\,+\,16b^2}{25} \\ =\,\frac{25a^2\,-\,9a^2\,+\,25b^2\,-\,16b^2\,-\,24ab}{25}\,=\,\frac{16a^2\,-\,24ab\,+\,9b^2}{25} \\ =\,$$\frac{4a\,-\,3b}{5}$$^2$ Since a and b are multiples of 5, (4a - 3b)/5 is always an integer, so a² + b² - c² is a perfect square.
 September 21st, 2013, 04:26 PM #3 Global Moderator   Joined: Dec 2006 Posts: 19,162 Thanks: 1638 It is not necessary that a and b are multiples of 5, but it's easy to see that 4a - 3b is a multiple of 5.
 September 21st, 2013, 05:17 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond Re: A really hard math problem "Prove that if..." Thanks for the correction. Can you explain why 4a - 3b is a multiple of 5?
 September 21st, 2013, 05:44 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,162 Thanks: 1638 You showed that ((4a - 3b)/5)² = a² + b² - c² (an integer). Alternative ending: a² + b² - c² ? (2a + b - 2c)² - (a - c)(3a + 4b - 5c) = (2a + b - 2c)².
 September 21st, 2013, 05:48 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond Re: A really hard math problem "Prove that if..." Thanks.
 September 22nd, 2013, 03:21 AM #7 Newbie   Joined: Sep 2013 Posts: 5 Thanks: 0 Re: A really hard math problem "Prove that if..." Thanks Guys @skipjack I do not fully understand your alternative ending; can you explain me that?
 September 22nd, 2013, 09:04 AM #8 Global Moderator   Joined: Dec 2006 Posts: 19,162 Thanks: 1638 a² + b² - c² ? (2a + b - 2c)² - (a - c)(3a + 4b - 5c) holds for any values of a, b and c. The right-hand side reduces to its first term, which is a perfect square, because the equation given in the problem can be simplified to 3a + 4b - 5c = 0, which implies that the second term is zero.
 September 22nd, 2013, 09:35 AM #9 Newbie   Joined: Sep 2013 Posts: 5 Thanks: 0 Re: A really hard math problem "Prove that if..." Oh Yeah, thanks again I invite you to think on my next problem viewtopic.php?f=13&t=42936

 Tags hard, math, problem, prove that if

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Tzad Algebra 14 September 14th, 2013 06:34 PM SedaKhold Calculus 0 February 13th, 2012 11:45 AM The Chaz Calculus 1 August 5th, 2011 09:03 PM katie0127 Advanced Statistics 0 December 3rd, 2008 01:54 PM muhammadmasood Abstract Algebra 6 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top