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If $4\mid a+bc$ and $6\mid b+ac$ prove that $2\mid a^2-b^2$

This is as far as I get:

$$a + bc = 4k\qquad\text{for some $k\in\Bbb Z$} \\b+ac =6l\qquad\text{for some $l\in\Bbb Z$}\\\implies (a^2-b^2)(1-c^2) = 16k^2 + 36l^2$$

Dave
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Lmorj
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    How did you get to your arrow? The algebra doesn't seem correct there. Instead, try and multiply your first equation by $a$, and your second by $b$... – Steven Stadnicki Sep 24 '17 at 02:15
  • @StevenStadnicki The algebra in fact checks out, apart from the fact that the right hand side should be $16k^2-36l^2$. The result in the body of the question was obtained from $(a+bc)^2-(b+ac)^2=(4k)^2-(6l)^2$. However, I agree that multiplying the first two equations by $a$ and $b$, respectively, is the better solution. See Math Lover's solution. – Dave Sep 24 '17 at 02:26

4 Answers4

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Following your approach, $$a + bc = 4k \implies a^2 + abc = 4ak,\tag 1$$ and $$b+ac =6l \implies b^2 + abc = 6bl. \tag 2$$ Subtract $(2)$ from $(1)$ to arrive at $$a^2-b^2 = 4ak-6bl = 2(2ak-3bl).$$ Q.E.D.

Math Lover
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Sometimes it pays off to think in much simpler terms, just involving parity. If $a$ is even, then so is $b$, because $b + ac$ is even. If $a$ is odd, then so are both $b$ and $c$, because $a + bc$ is even.

So $a$ and $b$ have the same parity, from which the result follows.

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If $4\mid a+bc$ and $6\mid b+ac$ prove that $2\mid a^2-b^2$

This is as far as I get:

$$a + bc = 4k\qquad\text{for some $k\in\Bbb Z$} \\b+ac =6l\qquad\text{for some $l\in\Bbb Z$}\\\implies (a^2-b^2)(1-c^2) = 16k^2 + 36l^2$$

Continue:

So $4|(a^2-b^2)(1-c^2)\;\;\;\;(*)$

If $c$ is odd, then $1-c^2 = 1-1-4k-4k^2 = -4k(k+1)$ and it is impossible to say that $2|a^2-b^2$. From here you must return to beginning formulas and do the thinking like user296602 did or something else.

If $c$ is even then $(1-c^2)$ is odd and thus the conclusion, but you really don't need this $(*)$ because you can again from starting formulas already see that $a$ and $b$ are even and you are done.

nonuser
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A variant of user296602's approach: If $c$ is even, then $a$ and $b$ must both be even; if $c$ is odd, then $a$ and $b$ must have the same parity. In either case $a^2-b^2$ is even.

Barry Cipra
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