# Find two consecutive odd integers such that the sum of their squares is 650?

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• Let 2n-1 and 2n+1 be the integers.

(2n-1)^2+(2n+1)^2 = 650

4n^2-4n+1+4n^2+4n+1=650

8n^2+2=650

8n^2=648

n^2=81

n= +/-9

If n=9 2n-1=17 and 2n+1=19

If n=-9 2n-1=-19 and 2n+1=-17

The numbers are 17 and 19 or -19 and -17.

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• :-

( 2n + 1 )² + ( 2n + 3 )² = 650

4n² + 4n + 1 + 4n² + 12n + 9 = 650

8n² + 16 n - 640 = 0

n² + 2n - 80 = 0

( n + 10 ) ( n - 8 ) = 0

n = - 10 , n = 8

One answer is 17 , 19

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• The numbers requied are 17 and 19

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• A=B+2,  A^2+B^2=650

A=-17 B=-19

A=19, B=17

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• 2n – 1, 2n + 1 are consecutive odd integers.

(2n – 1)² +  (2n + 1)²

= 8n² + 2 = 650

8n² = 648

n² = 81

n = ± 9

The numbers are (17, 19) and (-19, -17).

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• To make this simpler to solve, let's focus on the even integer that is between. Call this x.

Let x-1 be the odd integer immediately before.

Let x+1 be the odd integer immediately after.

The sum of their squares is 650:

(x - 1)² + (x + 1)² = 650

x² - 2x + 1 + x² + 2x + 1 = 650

2x² + 2 = 650

2x² = 648

x² = 324

x = ±√324

x = ±18

So the even number between is either -18 or 18.

-19 and -17

or

17 and 19

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• Half of 650 is 325, so the numbers should be near 18 (the square root of 324).

17^2 + 19^2 = 289 + 361 = 650.

No need to hunt any farther.

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• Two consecutive odd integers such that the sum of their squares is 650:

x^2 + (x + 2)^2 - 650 = 0

2x^2 + 4x - 646 = 0

x^2 + 2x - 323 = 0

Factorization:

(x + 19)(x − 17) = 0

17 and 19 & -17 and -19.

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• n^2+(n+2)^2 = 650

2n^2+4n+4=650

2n^+4n-646=0

n=17=> 17 and 19

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• x^2 + (x + 2)^2 = 650

x^2 + x^2 + 4x + 4 = 650

2x^2 + 4x + 4 = 650

x^2 + 2x + 2 = 325

x^2 + 2x + 1 = 324

(x + 1)^2 = 324

x + 1 = -18 , 18

x = -1 - 18 , -1 + 18

x = -19 , 17

-19 and -17

or

17 and 19

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