Question

If n is odd number and n > 1.then prove that (n, n²-1/2,n²+1/2)is a Pythagorean triplet.

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Answer 1

Among n, n² - ½, and n² + ½

n² + ½ is the biggest

Let a = n

b = n² - ½

c = n² + ½

c² = (n² + ½)²

c² = n⁴ + n² + ¼

||y

b² = n⁴ - n² + 1/4

a² = n²

a² + b² = n² + n⁴ - n² + ¼

a² + b² = ¼ - n⁴ ≠ c²

(???????)

(THERE MIGHT BE AN AMBIGUITY IN THE QUESTION, PLEASE RECHECK)

(imo the numbers are n, n-½ and n+½)

Answer 2

Answer:

Step-by-step explanation:

n2=(n2-1/2)^2+(n^2+1/2)^2

={(2n^2-1/2)}^2+{(2n^2+1/2)}^2

={[(2n^2)^2+(1)^2-2×2n^2×1]/4+{(2n^2)^2+(1)^2+2×2n^2×1)/4}

=(4n^4+1-4n^2)/4+(4n^4+1+4n^2)/4

=1/4(8n^4+2)

=(4n^4+1)/2