Elements of Algebra

CHAP. III. OF ALGEBRA. 81 $a+b$ $a-b$ --- $a^2+ab$ $-ab-b^2$ --- Product $a^2-b^2$. 276. Now, we may substitute for $a$ and $b$ any numbers whatever; so that the above example will furnish the following theorem; viz. The sum of two numbers, multiplied by their difference, is equal to the difference of the squares of those numbers: which theorem may be expressed thus: $(a+b) \times (a-b) = a^2-b^2$. And from this another theorem may be derived; namely, The difference of two square numbers is always a product, and divisible both by the sum and by the difference of the roots of those two squares; consequently, the difference of two squares can never be a prime number*. 277. Let us now calculate some other examples: $2a-3$ $a+2$ --- $2a^2-3a$ $4a-6$ --- $2a^2+a-6$ $4a^2-6a+9$ $2a+3$ --- $8a^3-12a^2+18a$ $12a^2-18a+27$ --- $8a^3+27$ $3a^2-2ab$ $2a-4b$ --- $6a^3-4a^2b$ $-12a^2b+8ab^2$ --- $6a^3-16a^2b+8ab^2$ $a^3+ab^3$ $a^3-a^3b^3$ --- $a^6+a^4b^3$ $-a^3b^3-a^4b^6$ --- $a^6-a^4b^6$ * This theorem is general, except when the difference of the two numbers is only 1, and their sum is a prime; then it is evident that the difference of the two squares will also be a prime: thus, $6^2-5^2=11$, $7^2-6^2=13$, $9^2-8^2=17$, \ldots.