For homework, I have to list all of the Pythagorean triples that include the number 24. Ex. 7,24,25... 24,45,51... If anyone knows a formula to answer this or someone that can help me please let me know. Thanks :)
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so a Pythagorean triple is in the form:
a=m^2 - n^2
b = 2mn
c = m^2 + n^2
with m>n>0 and m and n both being integers, we can then find every solution of these equations when a, b or c is equal to 24
I will start by solving for b=24, so
24 = 2mn
12=mn
because we restrict ourselves to integers and knowing m>n we get three possible solutions
m=12 n=1
m=6 n = 2
m=4 n=3
Now I will solve it for a = 24
24 = m^2 - n^2
m^2 = 24 + n^2
How I found the solutions to this is by knowing that (x+1)^2-x^2 = 2x+1, or the difference between the squares of two consecutive numbers is not only an odd number, but the nth positive odd number after 1, the difference here is even, so it must be the sum of an even number of positive odd consecutive integers, so I used this to find two more solutions(24/2 = 12 so 11+13, the smallest of these odds is 11 so (11-1)/2 = 5and two consecutive odds added so 7,5 and 24/4= 6 so 3+5+7+9=24 the smallest of these odds is 3, so (3-1)/2 = 1 and with 4 odd number so 5,1)
m=7 n = 5
n=5 n =1
There are no solutions for c = 24
The logic for this is because m^2 and n^2 are both positive and sum to 24 then m^2 and n^2 must be less than 24 and 24 is less then 25, so m and n must be less then 5, because 24 is even and we are restricted to integers we must either sum two odd or even number, because m>n they can not be equal, so we have two possibilities n=1, m=3 and n=2 m=4. Plugging these in we find they do not work (1^2 +3^2 = 10 and 2^2 + 4^2 = 20)
so we have 5 things to plug in so
m=12 n=1
m=6 n=2
m=4 n=3
m=5 n=1
m=7 n=5
so we get these 5 Pythagorean triples
143, 24, 145
32, 24,40
7,24,25
24,10,26
24,70,74
a=m^2 - n^2
b = 2mn
c = m^2 + n^2
with m>n>0 and m and n both being integers, we can then find every solution of these equations when a, b or c is equal to 24
I will start by solving for b=24, so
24 = 2mn
12=mn
because we restrict ourselves to integers and knowing m>n we get three possible solutions
m=12 n=1
m=6 n = 2
m=4 n=3
Now I will solve it for a = 24
24 = m^2 - n^2
m^2 = 24 + n^2
How I found the solutions to this is by knowing that (x+1)^2-x^2 = 2x+1, or the difference between the squares of two consecutive numbers is not only an odd number, but the nth positive odd number after 1, the difference here is even, so it must be the sum of an even number of positive odd consecutive integers, so I used this to find two more solutions(24/2 = 12 so 11+13, the smallest of these odds is 11 so (11-1)/2 = 5and two consecutive odds added so 7,5 and 24/4= 6 so 3+5+7+9=24 the smallest of these odds is 3, so (3-1)/2 = 1 and with 4 odd number so 5,1)
m=7 n = 5
n=5 n =1
There are no solutions for c = 24
The logic for this is because m^2 and n^2 are both positive and sum to 24 then m^2 and n^2 must be less than 24 and 24 is less then 25, so m and n must be less then 5, because 24 is even and we are restricted to integers we must either sum two odd or even number, because m>n they can not be equal, so we have two possibilities n=1, m=3 and n=2 m=4. Plugging these in we find they do not work (1^2 +3^2 = 10 and 2^2 + 4^2 = 20)
so we have 5 things to plug in so
m=12 n=1
m=6 n=2
m=4 n=3
m=5 n=1
m=7 n=5
so we get these 5 Pythagorean triples
143, 24, 145
32, 24,40
7,24,25
24,10,26
24,70,74