Given that
a^2 + b^2 + 9ab
is divisible by 11 (a and b are whole numbers), prove that
a^2  b^2
is also divisible by 11.
Divisible by 11

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Assume X = a  b
By substitution:
a^2 + (a + X)^2 + 9a(a +x)
Or:
a^2 + (a^2 + 2aX + X^2) + (9a^2 + 9aX)
Rearranging:
11a^2 + 11aX + X^2
The above equation will only be divisible by 11 when X^2 is divisible by 11. Which means X must be a factor of 11.
Factoring the second equation:
a^2  b^2 = (a+b)(ab) = (a+b)X
which will always be divisible by 11 when X is a factor of 11.
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Assume X = a  b
By substitution:
a^2 + (a + X)^2 + 9a(a +x)
Or:
a^2 + (a^2 + 2aX + X^2) + (9a^2 + 9aX)
Rearranging:
11a^2 + 11aX + X^2
The above equation will only be divisible by 11 when X^2 is divisible by 11. Which means X must be a factor of 11.
Factoring the second equation:
a^2  b^2 = (a+b)(ab) = (a+b)X
which will always be divisible by 11 when X is a factor of 11.
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Excellent!Whim wrote:<table bgcolor="white"><tr><td>
Assume X = a  b
By substitution:
a^2 + (a + X)^2 + 9a(a +x)
Or:
a^2 + (a^2 + 2aX + X^2) + (9a^2 + 9aX)
Rearranging:
11a^2 + 11aX + X^2
The above equation will only be divisible by 11 when X^2 is divisible by 11. Which means X must be a factor of 11.
Factoring the second equation:
a^2  b^2 = (a+b)(ab) = (a+b)X
which will always be divisible by 11 when X is a factor of 11.
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I'd just like to take a bit of issue with your "factor of 11" bit. 11 is a prime. It has no factors other than 1 and 11. It would be more accurate to say "is divisible by 11".
Anyway, great job!

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I noticed that too.DanishDynamite wrote:Just had another look at your proof, Whim.
You say X = a  b. You then substitute b in the original equation with (a + X). Somethings wrong with this picture.
X is defined to be a  b, but the equations use X as if it were equal to b  a.
This doesn't really change much; the factoring of a^2 + b^2 + 9ab comes out the same, but the factoring of a^2  b^2 comes out to be (a+b)(X), which still leads to essentially the same conclusion.

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Yes, the rest works out.Whim wrote:Sorry, my algebra is rusty. (Or maybe it's just my brain in general that is rusty.)
Let's change that to X = b  a.
That will make b = a + X. The rest should work out the same, yes?
Just one last quible before I show my proof. When you said...
...you didn't say why this is true. Just because X^2 is divisible by N doesn't mean X is divisible by N. For example, 6^2 = 36 is divisible by 18, but 6 is not divisble by 18. In other words, you forgot to mention that your deduction was true because 11 was a prime. :)The above equation will only be divisible by 11 when X^2 is divisible by 11. Which means X must be [divisible by] 11.
Anyway, here is the way I solved it:
Eq1 = a^2 + b^2 + 9ab
= a^2 + b^2 + 11ab  2ab
= a^2  2ab + b^2 + 11ab
= (ab)^2 + 11ab
Since we know Eq1 is divisible by 11, the above shows that (ab)^2 must also be divisible by 11. As 11 is a prime, we can deduce that (ab) is also divisible by 11.
The other equation which we need to show is divisible by 11 is:
Eq2 = a^2  b^2
= (a+b)*(ab)
Since (ab) is divisible by 11 so is Eq2.
QED.