Answer:
For any natural number m, we know that
2m, \mathrm{m}^{2}-1, \mathrm{m}^{2}+1m
2
−1,m
2
+1 is a Pythagorean triplet.
i. 2m = 6
\begin{array}{l} \Rightarrow \mathrm{m}=\frac{6}{2}=3 \\ \mathrm{m}^{2}-1=3^{2}-1=9-1=8 \\ \mathrm{m}^{2}+1=3^{2}+1=9+1=10 \end{array}
⇒m=
2
6
=3
m
2
−1=3
2
−1=9−1=8
m
2
+1=3
2
+1=9+1=10
∴ (6, 8, 10) is a Pythagorean triplet.
ii. 2m = 14
\begin{array}{l} \Rightarrow \mathrm{m}=\frac{14}{2}=7 \\ \mathrm{m}^{2}-1=7^{2}-1=49-1=48 \\ \mathrm{m}^{2}+1=7^{2}+1=49+1=50 \end{array}
⇒m=
2
14
=7
m
2
−1=7
2
−1=49−1=48
m
2
+1=7
2
+1=49+1=50
∴ (14, 48, 50) is not a Pythagorean triplet.
iii. 2m = 16
\begin{array}{l} \Rightarrow \mathrm{m}=\frac{16}{2}=8 \\ \mathrm{m}^{2}-1=8^{2}-1=64-1=63 \\ \mathrm{m}^{2}+1=8^{2}+1=64+1=65 \end{array}
⇒m=
2
16
=8
m
2
−1=8
2
−1=64−1=63
m
2
+1=8
2
+1=64+1=65
∴ (16, 63, 65) is a Pythagorean triplet.
iv. 2m = 18
\begin{array}{l} \Rightarrow \mathrm{m}=\frac{18}{2}=9 \\ \mathrm{m}^{2-1}=9^{2}-1=81-1=80 \\ \mathrm{m}^{2}+1=9^{2}+1=81+1=82 \end{array}
⇒m=
2
18
=9
m
2−1
=9
2
−1=81−1=80
m
2
+1=9
2
+1=81+1=82
∴ (18, 80, 82) is a Pythagorean triplet
Step-by-step explanation:
For any natural number greater than 1,(2m,m
2
−1,m
2
+1) is Pythagorean triplets.
So, if one number is 2m, then another two numbers will be m
2
−1 and m
2
+1.
Given, one number = 4
Then Pythagorean triplets:
2m=4 or m = 2
So,
m
2
−1=(2)
2
−1=4−1=3
m
2
+1=(2)
2
+1=4+1=5
Now, (3)
2
+(4)
2
=(5)
2
Or 9 + 16 = 25.
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