In a Pythagorean-like property, for any natural number m 1, the numbers 2m, m^2 - 1, m^2 + 1 form a triple where (2m)^2 + (m^2-1)^2 = (m^2+1)^2. For m = 4, find the value of m^2 + 1.

Question

In a Pythagorean-like property, for any natural number m > 1, the numbers 2m, m^2 - 1, m^2 + 1 form a triple where (2m)^2 + (m^2-1)^2 = (m^2+1)^2. For m = 4, find the value of m^2 + 1.

Accepted Answer

17. For m = 4, m^2 + 1 = 16 + 1 = 17. (The triple is 8, 15, 17 and 8^2 + 15^2 = 64 + 225 = 289 = 17^2.)

In a Pythagorean-like property, for any natural number $m > 1$ , the numbers 2 $m, m^{2} - 1, m^{2} + 1$ form a triple where $(2 m)^{2} + (m^{2} - 1)^{2} = (m^{2} + 1)^{2}$ .
For $m = 4$ , find the value of $m^{2} + 1$ .

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More Class 8 Properties of Square Numbers practice

In a Pythagorean-like property, for any natural number m>1, the numbers 2m,m2−1,m2+1 form a triple where (2m)2+(m2−1)2=(m2+1)2.For m=4, find the value of m2+1.

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 8 Properties of Square Numbers practice

In a Pythagorean-like property, for any natural number $m > 1$ , the numbers 2 $m, m^{2} - 1, m^{2} + 1$ form a triple where $(2 m)^{2} + (m^{2} - 1)^{2} = (m^{2} + 1)^{2}$ .
For $m = 4$ , find the value of $m^{2} + 1$ .