# Tag Info

Accepted

### 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 $$a+b+c\equiv a^2+b^2+c^2\equiv 2c^2\equiv 0\pmod 2$$
• 23.3k
Accepted

### 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 ...
• 2,781
Accepted

• 63.5k

### 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 ...
• 4,649
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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 $2 = \left(\frac{c}{a}\right)^2$. ...
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### 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.
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### 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 ...
• 376k
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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 ...
• 140k

### 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 ...
• 2,194

### 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 ...
• 63.5k
Accepted

### 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 ...
• 63.5k

### 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 $$a^2+b^2=c^2\tag{1}$$...
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### 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 ...
• 2,431
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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, ...
• 36.9k
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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 $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$.
• 4,028