SOLUTION: THe sum of the squares of three consecutive positve integers is equal to the sum of the squares of the next two integers. FInd the five integers.

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Question 187552: THe sum of the squares of three consecutive positve integers is equal to the sum of the squares of the next two integers. FInd the five integers.
Found 2 solutions by solver91311, jojo14344:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Let the first integer be x, then the second is x + 1, the third is x + 2, and so on.













or



Exclude the negative value because we are looking for positive integers. So the first integer is 10 and the others are therefore 11, 12, 13, 14.

Check:



You can do your own arithmetic to verify.

John

Answer by jojo14344(1513) About Me  (Show Source):
You can put this solution on YOUR website!


First Condition: "Sum of three consecutive positive integers"
x%5E2%2B%28x%2B1%29%5E2%2B%28x%2B2%29%5E2
Second Condition: "Equal to the sum of the next two integers"
%28x%2B3%29%5E2%2B%28x%2B4%29%5E2

Equating the two conditions being equal:
x%5E2%2B%28x%2B1%29%5E2%2B%28x%2B2%29%5E2=%28x%2B3%29%5E2%2B%28x%2B4%29%5E2, Equation 1
Expand:
x%5E2%2Bx%5E2%2B2x%2B1%2Bx%5E2%2B4x%2B4=x%5E2%2B6x%2B9%2Bx%5E2%2B8x%2B16
Addition/Subtraction (Cancellations):

Combine terms:
x%5E2%2B6x-14x%2B5-25=0---->x%5E2-8x-20=0
Factorable being perfect square:--->(x-10)(x+2)=0
Use highlighted, x=10, 1st Positive Integer, and follows the next four= 11,12,13,14
Go back Eqn 1 to verify:
10%5E2%2B11%5E2%2B12%5E2=13%5E2%2B14%5E2
100%2B121%2B144=169%2B196
365=365
Thank you,
Jojo