1. ## Positive integer solutions

Find the total number of positive integer solutions of:

2. $\displaystyle (x-y)(x+y)=33$

Then $\displaystyle \left\{\begin{array}{ll}x-y=1\\x+y=33\end{array}\right.$

or $\displaystyle \left\{\begin{array}{ll}x-y=3\\x+y=11\end{array}\right.$

Now find x and y.

3. Originally Posted by red_dog
$\displaystyle (x-y)(x+y)=33$

Then $\displaystyle \left\{\begin{array}{ll}x-y=1\\x+y=33\end{array}\right.$

or $\displaystyle \left\{\begin{array}{ll}x-y=3\\x+y=11\end{array}\right.$

Now find x and y.
Hello,Thank you,
Can not have :

BECAUSE ????????

4. Originally Posted by dhiab
Find the total number of positive integer solutions of:
Have you tried graphing the function?

You would probably have to enter it as

$\displaystyle y = \pm\sqrt{x^2 - 33}$.

I can see straight away that $\displaystyle x^2 - 33 \geq 0$ so $\displaystyle x \geq \sqrt{33}$.

Since $\displaystyle x$ needs to be a positive integer, we know therefore that $\displaystyle x \geq 6$.

Now you just need $\displaystyle x^2 - 33$ to be a perfect square. $\displaystyle x = 7$ works, since $\displaystyle 7^2 - 33 = 16$, so $\displaystyle y = 4$.

See if you can come up with any other solutions...

5. Originally Posted by Prove It
Have you tried graphing the function?

You would probably have to enter it as

$\displaystyle y = \pm\sqrt{x^2 - 33}$.

I can see straight away that $\displaystyle x^2 - 33 \geq 0$ so $\displaystyle x \geq \sqrt{33}$.

Since $\displaystyle x$ needs to be a positive integer, we know therefore that $\displaystyle x \geq 6$.

Now you just need $\displaystyle x^2 - 33$ to be a perfect square. $\displaystyle x = 7$ works, since $\displaystyle 7^2 - 33 = 16$, so $\displaystyle y = 4$.

See if you can come up with any other solutions...
Hello: the solutions is

In graphic of the function , we find the Point cordinates integer.
LOOK HERE

6. Originally Posted by dhiab
Hello: the solutions is

In graphic of the function , we find the Point cordinates integer.
LOOK HERE
Are there any more?

7. Originally Posted by Prove It
Are there any more?
Hello : yes in this question

Hello,Thank you,
Can not have :

BECAUSE

8. as the graph suggests, the values are symetric, mirror image of each other,
x = -/+ 17, y = -/+ 16
x = /+ 7 ; y -/+ 4

9. Hello, dhiab!

Here is a primitive solution . . .

Find the number of positive integer solutions of: .$\displaystyle x^2 - y^2 \:=\:33$
Consecutive squares differ by an odd number: .$\displaystyle (n+1)^2 - n^2 \:=\:2n+1$

We have: .$\displaystyle x^2 - y^2 \:=\:33$

. . Then: .$\displaystyle 2n+1 \:=\:33 \quad\Rightarrow\quad n \:=\:16$

. . Hence: .$\displaystyle \boxed{17^2 - 16^2 \:=\:33}$

$\displaystyle \text{We also have: }\;x^2-y^2 \:=\:\underbrace{9 +11+13}_{4^2\quad5^2\quad6^2\quad7^2}$

. . Hence: .$\displaystyle \boxed{7^2 - 4^2 \:=\:33}$

Therefore, there are exactly two solutions.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I came up with this procedure while in college.

Example: .$\displaystyle a^2-b^2 \:=\:105$

$\displaystyle 105$ is the difference between $\displaystyle \left(\tfrac{105-1}{2}\right)^2$ and $\displaystyle \left(\tfrac{105+1}{2}\right)^2$

. . Therefore: .$\displaystyle 53^2 - 52^2 \:=\:105$

We have: .$\displaystyle 105 \:=\:35 + 35 + 35 \:=\:33 + 35 + 37$

. . $\displaystyle \text{Hence: }\;a^2-b^2 \:=\:\underbrace{33 + 35 + 37}_{16^2\quad17^2\quad18^2\quad19^2}$

. . Therefore: .$\displaystyle 19^2 - 16^2 \:=\:105$

We have: .$\displaystyle 105 \:=\:15 + 15 + 15 + 15 + 15 + 15 + 15 \:=\:9 + 11 + 13 + 15 + 17 + 19 + 21$

. . $\displaystyle \text{Hence: }\;a^2-b^2 \:=\:\underbrace{9 + 11 + 13 + 15 + 17 + 19 + 21}_{4^2\quad\;\;5^2\quad\;\;6^2\quad\;\;\;7^2\qua d\;\;\;8^2\quad\;\;9^2\quad\;\;10^2\quad\;\;11^2}$

. . Therefore: .$\displaystyle 11^2 - 4^2 \:=\:105$

$\displaystyle \text{The next combination is: }\;105 \:=\:\underbrace{7+7+7+ \hdots + 7}_{\text{15 terms}}$

But this cannot be expressed as a sum of conscutive positive odd numbers.

Therefore, there are three solutions: .$\displaystyle (53,52),\;(19,16),\;(11,4)$

10. Originally Posted by Soroban
Hello, dhiab!

Here is a primitive solution . . .

Consecutive squares differ by an odd number: .$\displaystyle (n+1)^2 - n^2 \:=\:2n+1$

We have: .$\displaystyle x^2 - y^2 \:=\:33$

. . Then: .$\displaystyle 2n+1 \:=\:33 \quad\Rightarrow\quad n \:=\:16$

. . Hence: .$\displaystyle \boxed{17^2 - 16^2 \:=\:33}$

$\displaystyle \text{We also have: }\;x^2-y^2 \:=\:\underbrace{9 +11+13}_{4^2\quad5^2\quad6^2\quad7^2}$

. . Hence: .$\displaystyle \boxed{7^2 - 4^2 \:=\:33}$

Therefore, there are exactly two solutions.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I came up with this procedure while in college.

Example: .$\displaystyle a^2-b^2 \:=\:105$

$\displaystyle 105$ is the difference between $\displaystyle \left(\tfrac{105-1}{2}\right)^2$ and $\displaystyle \left(\tfrac{105+1}{2}\right)^2$

. . Therefore: .$\displaystyle 53^2 - 52^2 \:=\:105$

We have: .$\displaystyle 105 \:=\:35 + 35 + 35 \:=\:33 + 35 + 37$

. . $\displaystyle \text{Hence: }\;a^2-b^2 \:=\:\underbrace{33 + 35 + 37}_{16^2\quad17^2\quad18^2\quad19^2}$

. . Therefore: .$\displaystyle 19^2 - 16^2 \:=\:105$

We have: .$\displaystyle 105 \:=\:15 + 15 + 15 + 15 + 15 + 15 + 15 \:=\:9 + 11 + 13 + 15 + 17 + 19 + 21$

. . $\displaystyle \text{Hence: }\;a^2-b^2 \:=\:\underbrace{9 + 11 + 13 + 15 + 17 + 19 + 21}_{4^2\quad\;\;5^2\quad\;\;6^2\quad\;\;\;7^2\qua d\;\;\;8^2\quad\;\;9^2\quad\;\;10^2\quad\;\;11^2}$

. . Therefore: .$\displaystyle 11^2 - 4^2 \:=\:105$

$\displaystyle \text{The next combination is: }\;105 \:=\:\underbrace{7+7+7+ \hdots + 7}_{\text{15 terms}}$

But this cannot be expressed as a sum of conscutive positive odd numbers.

Therefore, there are three solutions: .$\displaystyle (53,52),\;(19,16),\;(11,4)$
Hello :
Thank you for your efforts, but génerale case is based on the divisores of a number