# PYTHAGORAS THEOREM!!!QUESTION 1: For what values of n does {n, n + 1, n + 2} form a pythagorean triple. QUESTION 2: Show that {n, n + 1, n + 3} cannot form a pythagorean triple.

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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Pythagoras theorem states that in a right angled triangle the square of hypotenuse equals sum squares of other two sides.

A Pythagoras triple refers to a set of three positive integers a,b, and c that satisfy the condition:

a^2 + b^2 = c^2

To establish that a set of three integers forms a Pythagoras Triple we have to prove that the sum of squares of the two smaller integers is equal to the square of the square of the third number.

Question 1

To find the value of n for which the set of numbers represented by (n, n+1, n+2) form a Pythagoras triple we form a equation based on on the condition of Pythagoras triple and then solve the equation for value of n.

Thus:

n^ + (n+1)^2 = (n+2)^2

n^ + (n+1)^2 - (n+2)^2 = 0

n^ + n^2 + 2n + 1- n^2 - 4n - 4 = 0

n^ - 2n - 3 = 0

n^ - 3n + n - 3 = 0

n(n - 3) + 1(n - 3) = 0

(n + 1)(n - 3) = 0

Therefore n = 3 0r n = -1

Thus the condition of Pythagoras triple is satisfied for n = 3

Question 2

To prove that the set of numbers represented by (n, n+1, n+3) cannot form a Pythagoras triple we form a equation based on on the condition of Pythagoras triple,  solve the equation for value of n, and then show that these values of n are not integers

Thus:

n^ + (n+1)^2 = (n+3)^2

n^ + (n+1)^2 - (n+3)^2 = 0

n^ + n^2 + 2n +1 - n^2 - 6n - 9 = 0

n^2 - 4n - 8 = 0

n^2 - 4n + 4 = 12

(n - 2)^2 = 12

n - 2 = 12^1/2 = 3.4641

n = 3.4641 + 2 = 5.4641

As only possible value of n is not an integer, the given set of number cannot form a Pythagoras triple.

neela | High School Teacher | (Level 3) Valedictorian

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Obviously the greatest side is the htpotenuse. So n+2 should be the sum of the squares on the other two sodes , n+1 and n forming the right angle. So,

(n+2)^2 = (n+1)^2+n^2. Or

(n^2+4n+4) = (n^2+2n+1)+n^2 . Or

0 = 2n2+2n+1 - (n^2+4n+4) . Or

0 = n^2-2n -3. Or

0 = (n+1)(n-3). Or

n-3 = 0 gives n =3. Therefore, 3 , 3+1= 4 and 3+2=5.

2)

Let us assume that n^2+(n+1)^2 =  (n+3)^2. Or

n^2 +n^2+2n+1= n^2+6n+9. Or

2n^2+2n- (n^2+6n ) = 9-1. Or

n^2-4n = 8 Or

n^2 -4n + 4 = 8+4 = 12 . Or

(n-2)^2 = sqrt12. Or

n = 2+2*3^(1/2).

Therefore , 2+2*3^(1/2), 3+2*3^(1/2)  and 5+2*3(1/2) are  the pytagorus triple, with the charecter, n, n+1 and n+3, but they are not the integers.