right triangle with prime sidelengths (2008 All-Siberian Qualifying 10.1)

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0 replies

jlacosta

Apr 2, 2025

0 replies

Eazy equation clap

giangtruong13 0

an hour ago

Find all $x,y,z$ satisfy that: $\frac{x}{y+z}=2x-1; \frac{y}{x+z}=3y-1;\frac{z}{x+y}=5x-1$

0 replies

giangtruong13

an hour ago

0 replies

inequality

revol_ufiaw 3

N 2 hours ago by MS_asdfgzxcvb

Prove that that for any real $x \ge 0$ and natural number $n$ ,
$x^n (n+1)^{n+1} \le n^n (x+1)^{n+1}.$

3 replies

revol_ufiaw

3 hours ago

MS_asdfgzxcvb

2 hours ago

How to judge a number is prime or not?

mingzhehu 0

3 hours ago

A=（10X1+1）(10X+1），X1，X∈N+
B=（10 X1+3）（10X+7），X∈N，X1∈N
C=(10 X1+9)(10X+9), X∈N，X1∈N
D=(10 X1+1)(10X+3), X1∈N+，X∈N
E=（10 X1+7）(10X+9），X∈N，X1∈N
F=（10 X1+1）（10X+7），X1∈N+，X∈N
G=（10 X1+3）（10X+9），X∈N，X1∈N
H=（10 X1+1）10X+9），X1∈N+，X∈N
I=（10 X1+3）（10X+3），X1∈N，X∈N
J=( 10X1+7)(10X+7),X∈N，X1∈N

For any natural number P∈{P=10N+1，n∈N}，make P=A or B or C
If P can make the roots of function group(ABC) without any root group completely made up of integer, P will be a prime
For any natural number P∈{P=10N+3，n∈N}，make P=D or E
If P can make the roots of function group(DE) without any root group completely made up
of integer, P will be a prime
For any natural number P∈{P=10N+7，n∈N}，make P=F or G
If P can make the roots of function group(FG) without any root group completely made up
of integer, P will be a prime
For any natural number P∈{P=10N+9，n∈N}，make P=H or I or J
If P can make the roots of function group(GIJ) without any root group completely made up
of integer, P will be a prime

0 replies

mingzhehu

3 hours ago

0 replies

What is an isogonal conjugate and why is it useful?

EaZ_Shadow 6

N 3 hours ago by maxamc

What is an isogonal conjugate and why is it useful? People use them in Olympiad geometry proofs but I don’t understand why and what is the purpose, as it complicates me because of me not understanding it.

6 replies

EaZ_Shadow

Dec 28, 2024

maxamc

3 hours ago

No more topics!

right triangle with prime sidelengths (2008 All-Siberian Qualifying 10.1)

parmenides51 3

N May 12, 2021 by jasperE3

Is there a right-angled triangle whose sidelengths are prime numbers?

3 replies

parmenides51

May 11, 2021

jasperE3

May 12, 2021

right triangle with prime sidelengths (2008 All-Siberian Qualifying 10.1)

G H J

Reveal topic tags

geometry

right triangle

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parmenides51

30629 posts

parmenides51

#1 h May 11, 2021, 10:53 PM

Y by

Is there a right-angled triangle whose sidelengths are prime numbers?

This post has been edited 1 time. Last edited by parmenides51, May 11, 2021, 11:03 PM
Reason: title typo

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OlympusHero

17019 posts

OlympusHero

#2 h May 11, 2021, 10:58 PM

Y by

Let no side length be equal to $2$ . Then the legs squared will each be odd, making their sum even, and the hypotenuse squared will be odd, contradiction. If the hypotenuse is $2$ , there are trivially no solutions. If a leg is $2$ , write $a^2+4=b^2$ . It is easy to see that there are no perfect squares differing by four that satisfy the prime condition, so the answer is no.

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rayfish

1121 posts

rayfish

#3 h May 11, 2021, 11:03 PM

Y by

Solution

The Answer is no. Consider a right triangle with integer side lengths $a$ , $b$ , and $c$ , where $c$ is the length of the hypotenuse.

We have three cases:
Case 1: $a$ and $b$ are even. This means $c$ must be even, which is impossible when $a$ , $b$ , and $c$ are prime.

Case 2: $a$ and $b$ are odd. This means $a^2$ and $b^2$ are congruent to $1$ modulo $4$ , so $c^2 \equiv 2 \pmod{4}$ , which is impossible if $c$ is a positive integer.

Case 3: One of $a$ or $b$ is even and the other is odd. The even leg must have length $2$ , meaning that the difference between the square of the other leg and the square of the hypotenuse is $4$ . Considering gaps between perfect squares, this has no solutions.

Thus, no triangle exists.

This post has been edited 1 time. Last edited by rayfish, May 11, 2021, 11:04 PM

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jasperE3

11165 posts

jasperE3

#4 h May 12, 2021, 1:17 AM

Y by

If $p,q,r$ are the sides, then $p^2+q^2=r^2$ . Looking at parity, we must have $2\in\{p,q,r\}$ , so WLOG $q=2$ since $r\ne2$ . Then $p^2=(r-2)(r+2)$ , so one of the following are true:
$r-2=1,p^2=r+2$
$r-2=p=r+2$
$r-2=p^2,r+2=1$
but these are all absurd, hence the answer is $\boxed{\text{no}}$ .

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