Another chance to brag about my latex!

Here is another one

To begin all word problems, we always have to turn the words into math.

Let's do this!

changing the words to math, we make an equation:

$$a^2-b^2=495$$

Now what do we do?

hmmmmm....do things seem familiar?

$$a^2-b^2=(a+b)(a-b)$$

Plugging that in the equation, we get:

$$(a+b)(a-b)=495$$

To try to plug all the factors in those parenthesis, we need to know the prime factorization.

$$495=3\cdot 165=3\cdot 5\cdot 33$$

(remember: 1 and 495 count as factors)

Lets first plug in 99 and 5.

$$(a+b)(a-b)=495$$

$$(99)(5)=495$$

which means $a+b=99$, and $a-b=5$

Now it's a system of equations!

$$a+b=99$$

$$a-b=5$$

Adding both equations:

$$2a=104$$

$$\rightarrow \div 2$$

$$a=52$$

Plugging $a$ in the equation $a+b=99$:

$$52+b=99$$

$$\rightarrow -52$$

$$b=47$$

To check that we did everything correctly, we can plug it in the other equation:

$$a-b=5$$

$$52-47=5$$

$$5=5$$

So we did everything correct!

It doesn't hurt if we check things twice! :) So lets do it.

$$a^2-b^2=495$$

$$52^2-47^2=495$$

(I used a calculator here :P)

$$2704-2209=495$$

$$495=495$$

Nice! We got one! Time to repeat the pattern now :(

Next, lets use the factors 15 and 33

$$(a+b)(a-b)=495$$

$$(15)(33)=495$$

$$a+b=15$$

$$a-b=33$$

adding the two equations:

$$2a=48$$

$$\rightarrow \div 2$$

$$a=24$$

plugging $a$ in:

$$24+b=15$$

$$\rightarrow -24$$

$$b=-9$$

plugging $a$ and $b$ in the other equations:

$$a-b=33$$

$$24--9=33$$

$$33=33$$

Checking again:

$$a^2-b^2=495$$

$$24^2--9^2=495$$

$$576-81=495$$

$$495=495$$

Time to work on the third pair of factors: 165, and 3

$$(a+b)(a-b)=495$$

$$(3)(165)=495$$

$$a+b=3$$

$$a-b=165$$

adding both equations:

$$2a=168$$

$$\rightarrow \div 2$$

$$a=84$$

plugging that in for the equation $a+b=3$

$$a+b=3$$

$$84+b=3$$

$$\rightarrow -84$$

$$b=-81$$

Plugging $a$ and $b$ in for the second equation:

$$a-b=165$$

$$84--81=165$$

$$165=165$$

Checking our work a second time:

$$a^2-b^2=495$$

$$84^2--81^2=495$$

$$7056-6561=495$$

$$495=495$$

Time to plug in our last factor: $1\cdot 495$

$$(a+b)(a-b)=495$$

$$(495)(1)=495$$(notice how I put the larger number in the $a+b$ parenthesis, so I don't have to deal with negatives.)

$$a+b=495$$

$$a-b=1$$

adding both equations:

$$2a=496$$

$$a=248$$

plugging in $a$:

$$248+b=495$$

$$\rightarrow -248$$

$$b=247$$

plugging in $a$ and $b$ for the second equation:

$$248-247=1$$

$$1=1$$

double-checking our work a second time:

$$a^2-b^2=495$$

$$248^2-247^2$$

$$61504-61009=495$$

$$495=495$$

SO ALL OF THE FACTORS WORK. WHICH GIVES US AN ANSWER OR $\boxed{4}$ possible squares that differ by 495.

Hmmmmmm. How come all of the factors are working? Something fishy is going on. :juggle: