what is the two pythagorean triples with 24 as one of the numbers?

Question

is it 7,24,25 and 18, 24, 30

Answer 1

Yes,
7² + 24² = 25² and 18² + 24² = 30²

Answer 2

Those are correct and so are
10,24,26; 24,70,74; 24, 32, 40 ; 24, 45, 51, and 24, 143, 145.
We will now show there are no others.
First, 24² = 576
and there are no positive integers, a, b, such that
a² + b² = 576.
Proof: Pick all squares less than 576 and check.
Or: Show that a and b must both be divisible by 3.
Set a = 3x, b = 3y.
Then x² + y² = 64.
Now check all squares less than 64 and show
that none yield a solution.
Thus 24 can never be the largest member of a
Pythagorean triple.
So now we may assume that 576 + b² = c².
This yields
c² - b² = 576
(c+b)(c-b) = 576.
So c-b and c+ b must be divisors of 576.
Further, since c and b are positive integers
and c = b = 24 does not yield a solution,
c+ b > 24, c - b < 24
Also the divisor pairs (576,1), (192, 3) and (64, 9)
do not yield integer solutions.
Now let's list each possible divisor pair
and find the solution.
1). (288,2).
So c+b = 288
c-b = 2
2c = 290 c = 145, b = 143.
Solution:
24, 143, 145.

2). (144,4)
c + b = 144
c -b = 4
c = 74, b= 70
Solution: 24,70, 74.

3). (96,6)
c + b = 96
c - b = 6
c = 51. b = 45
Solution: 24,45,51.

4). (72, 8)
c + b = 72
c - b =8
c = 40
b = 32
Solution: 24, 32, 40

5). (48,12)
c + b = 48
c-b = 12
c = 30, b = 18
Solution: 24, 18, 30, i.e., 18, 24, 30.

6). (36, 16)
c + b = 36
c - b = 16
c = 26, b = 10
Solution: 24, 10, 26, i.e., 10, 24,26.

7). (32, 18)
c + b = 32
c - b =18
c = 25, b = 7
Solution: 24, 7, 25, i.e., 7, 24, 25.

Since this exhausts all the possibilities, we
conclude that there are exactly 7 Pythagorean
triples with 24 as one of the numbers.
These are the 2 mentioned in the question
plus the 5 listed above.

Answer 3

All Pythagorean triples are generated by the formula
(k(a²-b²), 2kab, k(a²+b²)) i.e. k ((a²-b²), 2ab, (a²+b²))
where a>b ; also a and b have no common factor

Here are the cases people have found (listed by increasing k, which I think makes most sense)

7, 24 ,25 is the case k=1,a=4,b=3
24, 70, 74 is the case k=2,b=7,c=5
24, 45, 51 is the case k=3,a=4,b=1
10, 24, 26 is the case k=5,a=5,b=1
18, 24, 30 is the case k=6,a=2,b=1
24, 32, 40 is the case k=8,a=2,b=1
24, 143, 145 is the case k=24,a=12,b=1

Note the following test is order(sqrt(n)) which is more effivcient than steiner1574's brute force O(n) test.

To check whether these are the only ones:
Note the factorization of 24=2³3
And the factorizations of
(k(a²-b²), 2kab, k(a²+b²)) = (k(a-b)(a+b), 2kab, k(a²+b²))
So 24 can either be

Case a)
24=k(a-b)(a+b) =>(a²-b²)=1,2,3,4,6,12 or 24
Case b)
24=2kab =>ab=12=>k, a or b=1,2,3,4,6 or 12
Case c)
24=k(a²+b²) => (a²+b²)=1,2,3,4,6,12 or 24

so you just crunch these out and show the other ones do not have integer solutions.

Case a)
a²-b²=1,2,4,6 have no solution
a²-b²=(a-b)(a+b)=12 or 24
has solutions:
(1)(3) =>a=2,b=1 => (18,24,30) or (24,32,40)
(2)(6) => a=4,b=2 (invalid because have common factor 2)
(4)(6) => a=5,b=1 => (10, 24, 26)
(2)(12) => a=7,b=5 => (24,70,74)

Case c)
a²+b²=2 has nonsense solution a=1,b=1
a²+b²=1,3,4,6,12,24 all have no solution

Case b)
24=2kab =>ab=12=>k, a or b=1,2,3,4,6 or 12
remembering a>b, possible solutions are
k=1,a=4,b=3 => (7,24,25)
k=1,a=6,b=4 (invalid due to common factor 2)
k=1,a=12,b=1 => (24, 143, 145)
k=2,a=3,b=2 => (10,24,26)
k=2,a=6,b=1 => (24, 70, 74)
k=3,a=4,b=1 => (45,24,51)

Looks like all of the solutions.

Answer 4

ok Smile, that could be a chunk of cake. Pythagorean triples refers back to the pythagorean theorem. enable's ascertain that we are on the comparable web site... Pythagorean Theorem - assume you have a suited Triangle (one that has a suited attitude interior of it). The area opposite the main suitable attitude is the hypotenuse. enable the hypotenuse have a length 'C'. the different 2 aspects (adjoining to the main suitable attitude) are lengths 'A' and 'B'. Pythagorus suggested that: in case you have a suited triangle, Then : A^2 + B^2 = C^2 A PYTHAGOREAN TRIPLE is purely a series of numbers that fulfill the above circumstances. for occasion, the numbers 3,4 and 5 are a pythagorean triple ( 3^2 + 4^2 = 5^2). yet another triple is 5,12 and 13 (5^2 + 12^2 = 13^2). There are a limiteless variety of triples that fulfill pythagorean theorem, yet do not make the blunders of assuming ANY 3 numbers will. case in point, 3, 8 and 9 are actually not a triple (3^2 + 8^2 does not equivalent 9^2). Triples are specific circumstances that we sparkling up for using pythagorean theorem. Now, enable's do your issues... subject a million - One leg is 20 and the different is 15. The hypotenuse is 'C'. using pythagorean theorem all of us be attentive to : A^2 + B^2 = C^2 the place: A=20 , B=15 , and the hypotenuse is 'C'. Plugging those values in we detect... (20^2) + (15^2) = C^2 fixing for C... C = SQRT ( 20^2 + 15^2) C=25 subject 2 - One leg is 24 the different is 'x'. The hypotenuse is forty... A=24 , B=x and C=forty Plugging into pythagorean theorem... 24^2 +x^2 = forty^2 x^2 = forty^2 - 24^2 x^2 = 1024 x= 32 wish that facilitates. sturdy success!

Answer 5

Yes but.
7, 24, 30
7^2+24^2=49+576=625=25^2

18, 24, 30
18^2+24^2=324+576=900=30^2

BTW 18, 24, 30is just the 3, 4, 5 triple multiplied by 6.

AND
24, 143, 145
24^2+143^2=576+20449=21025=145^2

also

10, 24, 26
10^2+24^2=100+576=676=26^2

this is the 5, 12, 13 triple multiplied by 2.

My guess is that they want 7, 24, 25 & 24, 143, 145

Answer 6

What about 24-32-40?

If your question says there should be only two, maybe this one and 18-24-30 aren't allowed because they both reduce to 3-4-5...

Answer 7

10, 24, 26 and 18, 24, 30 and 7,24,25
There appears to be at least three.

Answer 8

There's actually 3:

7, 24, 25
18, 24, 30
10, 24, 26

Answer 9

There are several possibilities. The two you mention, plus:

10-24-26
24-32-40
24-45-51

Answer 10

yes it is 7,24,25 and 18,24,30