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Prove the statement- "For all integers a,b, and c if a∣b and a∣(b^2 −c), then a∣c.

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Answer:

To prove the statement "For all integers a, b, and c, if a divides b and a divides (b^2 - c), then a divides c," we will use the definition of divisibility.

Assuming a divides b, we can write b = ka, where k is an integer.

Similarly, assuming a divides (b^2 - c), we can write b^2 - c = la, where l is an integer.

We want to show that a divides c, meaning that c = ma, where m is an integer.

First, let's consider the expression (b^2 - c). We substitute b = ka into this expression:

(b^2 - c) = (ka)^2 - c

= k^2a^2 - c

Since a divides (b^2 - c), we can rewrite this as:

k^2a^2 - c = pa

where p is some integer.

Expanding the left side of the equation, we have:

k^2a^2 - c = pa

k^2a^2 = pa + c

k^2a^2 = a(p + ka)

Now we observe that a divides the left side of the equation, k^2a^2, and since a also divides the right side of the equation, a(p + ka), we can conclude that a divides c, or c = ma, where m = p + ka, an integer.

Therefore, if a divides b and a divides (b^2 - c), then a divides c, completing the proof.

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