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 Find 5 consecutive whole numbers if it is known that the sum of the squares of the first 3 numbers is equal to the sum of the squares of the last 2 numbers.

User Argoron
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1 Answer

2 votes
so... our numbers... let's say the first one is hmmm "a"
so the second and subsequent are
a
a+1
a+2
a+3
a+4

there, 5 consecutive whole numbers or integers for that matter

now, we know the sum of the square of the first three,
is the same as the sum of the square of the last two

so
\bf \begin{cases} a\\ a+1\\ a+2\\ \textendash\textendash\textendash\textendash\\ a+3\\ a+4 \end{cases}\qquad (a)^2+(a+1)^2+(a+2)^2=(a+3)^2+(a+4)^2

do a binomial theorem expansion on those, solve for "a"
User Kapitan
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