document.write( "Question 430334: let 'a' and 'b' be a pair of positive integers whose highest common factor is 1.
\n" ); document.write( "sometimes the sum of such pairs is even, sometimes it is odd.
\n" ); document.write( "consider just those pairs for which a+b is even. let 'x' = ab(a^2-b^2)
\n" ); document.write( "After trying a few cases you will probably be led to conjecture that 'x' is divisible by 24.
\n" ); document.write( "Determine if there exists any such 'x' which are not divisible by 24.
\n" ); document.write( "Note: determine in this case means to show evidence and working out \n" ); document.write( "
Algebra.Com's Answer #298852 by richard1234(7193)\"\" \"About 
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We have \"x+=+ab%28a%5E2+-+b%5E2%29+=+ab%28a-b%29%28a%2Bb%29\". If a+b is even, then a and b must both be odd (if they were both even then GCF(a,b) >= 2). Also, if a+b is even, then a-b is also even.\r
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\n" ); document.write( "\n" ); document.write( "We can try to find some ordered pair (a,b) such that a+b ≡ 2 (mod 4) and a-b ≡ 2(mod 4). However, this fails because a-b = (a+b) - 2b ≡ 2 - 2 ≡ 0 (modulo 4) (taking into account that b is even). This implies that if a-b ≡ 2 (mod 4), then a+b ≡ 0 (mod 4). Similarly, if a+b ≡ 2 (mod 4), then a-b ≡ 0 (mod 4). This means that for any (a,b), x ≡ 0 (mod 8).\r
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\n" ); document.write( "\n" ); document.write( "Next, we need to show whether x must be divisible by 3. We can assume that a,b ≡ 1 or 2 (mod 3). However, if a ≡ b (mod 3), then a-b ≡ 0 (mod 3), and if a ≡ 1, b ≡ 2, or if a ≡ 2, b ≡ 1 (mod 3), then a+b ≡ 0 (mod 3). In either case, x ≡ 0 (mod 3).\r
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\n" ); document.write( "\n" ); document.write( "We have shown that x ≡ 0 (mod 8) and x ≡ 0 (mod 3). Since 8 and 3 are relatively prime, then x ≡ 0 (mod 24) so for all integers a and b, x is divisible by 24. \n" ); document.write( "
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