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So I am given that $65 = 1^2 + 8^2 = 7^2 + 4^2$ , how can I use this observation to find two Pythagorean triangles with hypotenuse of 65.

I know that I need to find integers $a$ and $b$ such that $a^2 + b^2 = 65^2$, but I don't understand how to derive them from that observation.

Here is my attempt.

$65^2 = (8^2+1^2)(7^2+4^2) = 8^27^2 + 1^24^2 + 1^27^2 + 8^24^2 = (8\cdot7)^2 + 4^2 + 7^2 + (8\cdot4)^2$ but now I am stuck here, any suggestions!

Bill Dubuque
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alkabary
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  • use Euclid's formula – Asinomás Jan 16 '15 at 06:06
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    Another way is to derive the triple from more famous triples. With 65, for example, since $65 = 5 \times 13$, and since $3^2 + 4^2 = 5^2$ and $5^2 + 12^2 = 13^2$, we can multiple the first triple by 13 and the second triple by 5, giving us: $$39^2 + 52^2 = 65^2$$ and $$25^2 + 60^2 = 65^2.$$ – Robert Soupe Jan 16 '15 at 06:09
  • yes, but i am not using the observation here. Thanks though – alkabary Jan 16 '15 at 06:13
  • The formula there. http://math.stackexchange.com/questions/153603/diophantine-equation-a2b2-c2d2/736164#736164 – individ Jan 16 '15 at 06:30
  • If it is necessary to solve the system of the formula there. http://www.artofproblemsolving.com/blog/104785 – individ Jan 16 '15 at 06:35

4 Answers4

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Use the Brahmagupta-Fibonacci Identity $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2.$$ This identity can be verified by multiplying out each side, and in nicer ways.

From the Identity, we get $$65^2=(8^2+1^2)(7^2+4^2)=(8\cdot 7-1\cdot 4)^2 +(8\cdot 4+1\cdot 7)^2.$$

We can get another representation of $65^2$ as the sum of two squares by letting $c=4$ and $d=7$.

Remark: The Identity gives the useful result that the product of two numbers, each the sum of two squares, is itself the sum of two squares.

André Nicolas
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  • Much thanks, but this is one Pythagorean triple,How can i get another one form the same observation – alkabary Jan 16 '15 at 06:35
  • I added a line about how. – André Nicolas Jan 16 '15 at 06:41
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    @Leo: Maybe this will help you see how we can get two triples: $$\begin{align} \ (a^2 + b^2)(c^2 + d^2) & = \ & = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 \ & = a^2c^2 + b^2d^2 \pm 2acbd + a^2d^2 + b^2c^2 \mp 2adbc \ & = (ac \pm bd)^2 + (ad \mp bc)^2\ \end{align}$$ – PM 2Ring Jan 16 '15 at 06:44
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Hint If $m,n$ are integers then $(m^2-n^2, 2mn, m^2+n^2)$ is a Pytagorean triple.

N. S.
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As I said we need to solve a system of equations.

The system of equations:

$$\left\{\begin{aligned}&x^2+y^2=z^2\\&q^2+t^2=z^2\end{aligned}\right.$$

the solutions have the form: $$x=4p^4-s^4$$ $$y=4p^2s^2$$ $$q=4ps(2p^2-s^2)$$ $$t=4p^4-8p^2s^2+s^4$$ $$z=4p^4+s^4$$

$p,s,k$ - integers.

Formulas you can write a lot, but will be limited to this. Will make a replacement.

$$a=p^2+s^2-k^2$$

$$b=p^2+s^2+k^2-2pk-2ks$$

$$c=p^2+k^2-s^2+2ps-2kp$$

$$r=s^2+k^2-p^2+2ps-2ks$$

The solution then is.

$$x=2ab$$

$$y=a^2-b^2$$

$$q=2cr$$

$$t=c^2-r^2$$

$$z=a^2+b^2$$

individ
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If we solve $C=m^2+n^2$ for $n$, we get $n=\sqrt{C-m^2}$. We can then try values of $m$ to see which one(s) yield a positive integer for $n$, where

$$\text{To ensure }n<m:\biggl\lceil\sqrt{\frac{C}{2}}\space\space\biggr\rceil \le m\le\bigl\lfloor\sqrt{C}\bigr\rfloor:\text{ to ensure }n\in\mathbb{N}$$

$$\text{In the example of }65\qquad m_{min}=\lceil\sqrt{32.5}\space\rceil = 6\qquad m_{max}=\lfloor\sqrt{65}\rfloor = 8$$

If we test $n=\sqrt{C-m^2}$ using $m=6,7\text{, and }8$, we find positive integers for $n$ where $m=7\Rightarrow n=4$ and $m=8\Rightarrow n=1.$

$$\text{Using Euclid's formula }f(7,4)=(33,56,65)\quad f(8,1)=(63,16,65)$$

poetasis
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