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September 21st, 2013, 03:19 PM   #1
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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
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September 21st, 2013, 05:01 PM   #2
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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.





Since a and b are multiples of 5, (4a - 3b)/5 is always an integer, so a + b - c is a perfect square.
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September 21st, 2013, 05:26 PM   #3
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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.
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September 21st, 2013, 06:17 PM   #4
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Re: A really hard math problem "Prove that if..."

Thanks for the correction. Can you explain why 4a - 3b is a multiple of 5?
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September 21st, 2013, 06:44 PM   #5
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You showed that ((4a - 3b)/5) = a + b - c (an integer).

Alternative ending: a + b - c ? (2a + b - 2c) - (a - c)(3a + 4b - 5c) = (2a + b - 2c).
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September 21st, 2013, 06:48 PM   #6
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Re: A really hard math problem "Prove that if..."

Thanks.
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September 22nd, 2013, 04:21 AM   #7
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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?
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September 22nd, 2013, 10:04 AM   #8
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a + b - c ? (2a + b - 2c) - (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.
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September 22nd, 2013, 10:35 AM   #9
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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
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