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151 votes
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Prove that the sum of pythagorean triples is always even

Note that $x^2\equiv x\pmod 2$ and thus $a^2+b^2=c^2$ implies $$a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$$
Ennar's user avatar
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41 votes
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Is there a Pythagorean triple whose angles are 90, 45, and 45 degrees?

You cannot have an integer Pythagorean Triple whose angles are $45°, 45°$ and $90°$. Assume on the triangle we have sides $a$. Then by Pythagoras' Theorem, $$a^2+a^2=2a^2=(a\sqrt{2})^2$$ This ...
Landuros's user avatar
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35 votes
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A very different property of primitive Pythagorean triplets: Can number be in more than two of them?

Start from $1105 = 5\cdot 13\cdot 17$ whose prime factors all has the form $4k+1$. Decompose the prime factors in $\mathbb{Z}$ over gaussian integers $\mathbb{Z}[i]$ as $$5 = (2+i)(2-i),\quad 13 = (3+...
achille hui's user avatar
31 votes
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Can a right triangle have odd-length legs and even-length hypotenuse?

Hint: Suppose you have $A^2+B^2=C^2$ where $A=2a+1, B=2b+1, C=2c$ Substitute, expand and then take out terms with factors of $4$
Henry's user avatar
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30 votes
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Does any given integer only occur in one primitive Pythagorean triple?

Yes, a number can appear as the smallest value in two distinct primitive triples. For example, $(57, 176, 185)$ and $(57, 1624, 1625)$. In fact, choose any two relatively prime positive integers $p$ ...
rogerl's user avatar
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25 votes

Can a right triangle have odd-length legs and even-length hypotenuse?

Suppose that you are right and there is a possible right triangle such that $a=2k+1, b=2m+1$ and $c=2n$ {legs are odd while hypotenuse is even} and you have $a^2+b^2=c^2$. Now on expanding (...
Vidyanshu Mishra's user avatar
25 votes
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How do you find Pythagorean triples that approximately correspond to a right triangle with a given angle?

Let $r\in[0,\infty)$. The problem posed is equivalent to finding $m\geq n\in\mathbb N$ such that $r\sim\frac{2mn}{m^2-n^2}$. Thus, we want $$rm^2-2mn-rn^2\sim0$$Now suppose $r,n$ is given and we want ...
Rushabh Mehta's user avatar
20 votes

Prove that the sum of pythagorean triples is always even

Also, Pythagorean triples have a well defined structure: $$a=k(m^{2}-n^{2}),\ \,b=k(2mn),\ \,c=k(m^{2}+n^{2})$$ and $$a+b+c=2k(mn+m^2)$$
rtybase's user avatar
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20 votes

How does Wiles' proof fail at $n=2$?

tl;dr version: A solution $a^\ell + b^\ell = c^\ell$ would generate a weird (i.e,. not modular) elliptic curve over $\mathbb{Q}$ for prime $\ell \geq 5$. Wiles proved that sufficiently weird elliptic ...
anomaly's user avatar
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18 votes

Prove that the sum of pythagorean triples is always even

Hint Write $a+b+c=k$, so $$a^2+b^2=(a+b)^2-2ab= (k-c)^2-2ab=c^2 → k^2-2(kc+ab)=0→k^2=2(kc+ab)$$
Arnaldo's user avatar
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18 votes

Prove that the sum of pythagorean triples is always even

Notice that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=2c^2+2(ab+bc+ac)$, so the square of $a+b+c$ is even and thus $a+b+c$ is also even.
John Doe's user avatar
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17 votes
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How does Wiles' proof fail at $n=2$?

The other "accepted" answer in my opinion more or less misses the point. So here is another answer. Consider the elliptic curve $$E: y^2 = x(x - a^p)(x + b^p).$$ The $p$-torsion points $E[p]$...
user297024's user avatar
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16 votes

Does any given integer only occur in one primitive Pythagorean triple?

The example $(3,4,5)$ and $(5,12,13)$ shows that some positive integers can appear in more than one primitive triple.
carmichael561's user avatar
16 votes

A very different property of primitive Pythagorean triplets: Can number be in more than two of them?

The "smallest" counterexample is \begin{gather*} 5^2 + 12^2 = 13^2 \\ 9^2 + 12^2 = 15^2 \\ 12^2 + 16^2 = 20^2 \end{gather*} ($12$ also satisfies $12^2 + 35^2 = 37^2$.) Edit: now that the OP has ...
lokodiz's user avatar
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16 votes
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$x^2+y^2=z^n$: Find solutions without Pythagoras!

The solution $(x,y,z)=(0,1,1)$ works for all $n$. If you don't want to allow $0$, then let $x,y\in\Bbb{N}$ be such that $$x+yi=(1+2i)^n.$$ Then $$5^n=((1+2i)(1-2i))^n=(1+2i)^n(1-2i)^n=(x+yi)(x-yi)=x^...
Servaes's user avatar
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16 votes

Is $100$ the only square number of the form $a^b+b^a$?

$\newcommand{\eps}{\varepsilon}$ $\newcommand{\rad}{\mathrm{rad}}$ At least, under the abc conjecture, there can be only finitely many pairs $(a,b)$ with $b>a>1$ coprime such that $a^b+b^a$ is ...
W-t-P's user avatar
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15 votes
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Pythagorean Triple where $a=b$?

Yes, it does revolve around the fact that $\sqrt{2}$ is irrational. For if there were integers $a$ and $c$ such that $a^2 + a^2 = c^2$, then $2a^2 = c^2$, or $2 = \left(\frac{c}{a}\right)^2$. ...
Matthew Leingang's user avatar
15 votes

Is there a Pythagorean triple whose angles are 90, 45, and 45 degrees?

No, since if the perpendicular sides are $a$ in length, the hypotenuse would be $a\sqrt2$. But $\sqrt2$ is irrational, so $a\sqrt2$ is not an integer.
user_194421's user avatar
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15 votes

Is there a Pythagorean triple whose angles are 90, 45, and 45 degrees?

In your context you might be interested in isosceles triangles that are almost right. As others have said, a right isosceles triangle has sides that are $a,a,a\sqrt 2$ and as $\sqrt 2$ is not ...
Ross Millikan's user avatar
14 votes
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Why can no prime number appear as the length of a hypotenuse in more than one Pythagorean triangle?

This goes back to Euler, who showed that if there are two ways of writing an odd integer $N$ as the sum of two squares, then $N$ is composite. There is a 2009 article on this by Brillhart. Let me try ...
Will Jagy's user avatar
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14 votes

Why can no prime number appear as the length of a hypotenuse in more than one Pythagorean triangle?

As noted in the comments and the accepted answer, this comes down to the fact that if a prime $p$ can be written as a sum of two squares, then the representation is unique up to switching and or ...
Joe Silverman's user avatar
13 votes

Is $100$ the only square number of the form $a^b+b^a$?

(21-03-2020) Update: As no unconditional answer has yet been given, I will include a few more conditions that any solution must satisfy. The results used are rather advanced, so I will only include ...
Servaes's user avatar
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13 votes
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Why is the one quadratic polynomial a perfect square more often than the other?

I should start by clarifying that both equations yield the same number of squares; both yield countably infinitely many perfect squares. Up to any given upper bound, however, the former equation ...
Servaes's user avatar
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12 votes

the converse of Pythagoras Theorem

The user @MvG noted that it should be an answer rather than a comment. Let $\triangle ABC$ be some arbitrary triangle with sides $a,b,c$ (assume that $AB=a,BC=b,AC=c$) such that $$a^2+b^2=c^2\tag{1}$$...
Galc127's user avatar
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12 votes

A very different property of primitive Pythagorean triplets: Can number be in more than two of them?

$120,160,200$ and $90,120,150$ and $72,96,120$ How I arrived at it: It is a common knowledge that if we scale the triplet $3,4,5$ by any constant, we get another triplet. So I found out a common ...
Shraddheya Shendre's user avatar
12 votes
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How can you find a Pythagorean triple with $a^2+b^2=c^2$ and $a/b$ close to $5/7$?

Pythagorean triplets are characterized by being of the form $$a= m^2-n^2 \\ b= 2mn \\ c=m^2+n^2$$ So you could look for two integers $m>n$ such that $$\frac{m^2-n^2}{2mn} \approx \frac{5}{7}$$ or, ...
Crostul's user avatar
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12 votes
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If $a+b+c$ divides the product $abc$, then is $(a,b,c)$ a Pythagorean Triple?

You actually want it the other way around: if $a^2+b^2=c^2$ then $a+b+c|abc$. That you can prove very quickly from the general form of primitive Pythagorean triples $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$.
Michal Adamaszek's user avatar
12 votes
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Arithmetic progression $(a,b,c)$ with $a^2+b^2=c^2$ and $\gcd(a,b,c) = 1$

A Pythagorean triple in arithmetic progression would be $(a-d, a, a+d)$, where $$(a-d)^2+a^2=(a+d)^2.$$ $$a^2-2ad+d^2+a^2=a^2+2ad+d^2$$ $$a^2=4ad$$ Given $a>0$, we have $a=4d$. Thus, the triple is $...
J. W. Tanner's user avatar
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11 votes
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Solve Euler Project #9 only mathematically - Pythagorean triplet

Hint Euclid's parameterization of the Pythagorean triples (Elements, Book X, Proposition XXIX) is: $$a = k (m^2 - n^2), \qquad b = 2 k m n, \qquad c = k (m^2 + n^2),$$ where $m > n > 0$ and $m, ...
Travis Willse's user avatar

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