# Positive integer solutions

• July 1st 2009, 01:37 AM
dhiab
Positive integer solutions
Find the total number of positive integer solutions of:
http://www.mathramz.com/xyz/latexren...438956f719.png
• July 1st 2009, 01:47 AM
red_dog
$(x-y)(x+y)=33$

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

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

Now find x and y.
• July 2nd 2009, 12:37 AM
dhiab
Quote:

Originally Posted by red_dog
$(x-y)(x+y)=33$

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

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

Now find x and y.

Hello,Thank you,
Can not have :
http://www.mathramz.com/xyz/latexren...a784fe38a8.png
BECAUSE ????????
• July 2nd 2009, 12:43 AM
Prove It
Quote:

Originally Posted by dhiab
Find the total number of positive integer solutions of:
http://www.mathramz.com/xyz/latexren...438956f719.png

Have you tried graphing the function?

You would probably have to enter it as

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

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

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

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

See if you can come up with any other solutions...
• July 2nd 2009, 01:51 AM
dhiab
Quote:

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

You would probably have to enter it as

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

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

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

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

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

Hello: the solutions is
http://www.mathramz.com/xyz/latexren...c39bfaaa03.png
In graphic of the function http://www.mathramz.com/xyz/latexren...1c04c218bd.png , we find the Point cordinates integer.
LOOK HERE (Evilgrin)
• July 2nd 2009, 01:55 AM
Prove It
Quote:

Originally Posted by dhiab
Hello: the solutions is
http://www.mathramz.com/xyz/latexren...c39bfaaa03.png
In graphic of the function http://www.mathramz.com/xyz/latexren...1c04c218bd.png , we find the Point cordinates integer.
LOOK HERE (Evilgrin)

Are there any more?
• July 2nd 2009, 10:20 AM
dhiab
Quote:

Originally Posted by Prove It
Are there any more?

Hello : yes in this question

Hello,Thank you,
Can not have :
http://www.mathramz.com/xyz/latexren...a784fe38a8.png
BECAUSE

http://www.mathramz.com/xyz/latexren...8c8e7e9f76.png
• August 12th 2009, 03:51 AM
pacman
as the graph suggests, the values are symetric, mirror image of each other,
x = -/+ 17, y = -/+ 16
x = /+ 7 ; y -/+ 4
• August 12th 2009, 06:57 AM
Soroban
Hello, dhiab!

Here is a primitive solution . . .

Quote:

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

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

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

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

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

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

Therefore, there are exactly two solutions.

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

I came up with this procedure while in college.

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

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

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

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

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

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

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

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

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

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

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

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

• August 12th 2009, 10:30 AM
dhiab
Quote:

Originally Posted by Soroban
Hello, dhiab!

Here is a primitive solution . . .

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

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

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

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

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

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

Therefore, there are exactly two solutions.

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

I came up with this procedure while in college.

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

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

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

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

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

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

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

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

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

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

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

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

Hello :
Thank you for your efforts, but génerale case is based on the divisores of a number