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# How can the fact that 24^2-5^2=551 be used to find the factors of 551 (not including 1 or 551)?

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How can the fact that 24^2-5^2=551 be used to find the factors of 551 (not including 1 or 551)?

Aug 8, 2017

#1
+1

The difference between the two numbers is always a factor of the difference between their  squares:
So, you have: 24 - 5 = 19 must be a factor of 551. And the sum of the two numbers is the other factor of the difference between their squares: So, you have: 24 + 5 =29 is the other factor of 551.
So, 551 = 19 x 29. This holds true for any two numbers and the difference between their squares.
Another Example: 67^2 - 22^2 =4,005. And 4,005=[67 - 22] x [67 + 22].
Note: The two factors are not ALWAYS prime factors, but two DIVISORS, which can be factored into their prime factors. So [67 - 22]=45 =3 x 3 x 5, and [67 + 22]=89, which is prime number in  this example. So, 4,005 =3^2 x 5 x 89.

Aug 9, 2017

#1
+1

The difference between the two numbers is always a factor of the difference between their  squares:
So, you have: 24 - 5 = 19 must be a factor of 551. And the sum of the two numbers is the other factor of the difference between their squares: So, you have: 24 + 5 =29 is the other factor of 551.
So, 551 = 19 x 29. This holds true for any two numbers and the difference between their squares.
Another Example: 67^2 - 22^2 =4,005. And 4,005=[67 - 22] x [67 + 22].
Note: The two factors are not ALWAYS prime factors, but two DIVISORS, which can be factored into their prime factors. So [67 - 22]=45 =3 x 3 x 5, and [67 + 22]=89, which is prime number in  this example. So, 4,005 =3^2 x 5 x 89.

Guest Aug 9, 2017
#2
+21848
+1

How can the fact that 24^2-5^2=551 be used to find the factors of 551 (not including 1 or 551)?

$$\begin{array}{|rcll|} \hline \text{Binomial Theorem} \\ (a+b)^2 &=& a^2+2ab+b^2 \\ (a-b)^2 &=& a^2-2ab+b^2 \\ (a-b)(a+b) &=& a^2-b^2 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline (a-b)(a+b) &=& a^2-b^2 \quad & | \quad a = 24 \qquad b = 5 \\ (24-5)(24+5) &=& 24^2-5^2 \\ 19\cdot 29 &=& 24^2-5^2 \\ \mathbf{19\cdot 29} & \mathbf{=} & \mathbf{551} \\ \hline \end{array}$$

Aug 9, 2017