4. Find all triples of consecutive numbers ,whose sum of squares is equal to some fourdigit number with all four digits being equal.
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22.06.2017 01:04
XbenX wrote: 4. Find all triples of consecutive numbers ,whose sum of squares is equal to some fourdigit number with all four digits being equal. Do you mean "Find all triples of consecutive numbers where the sum of the squares of the numbers is equal to some four digit number with all four digits being equal"?
22.06.2017 01:32
23.06.2017 02:22
26.04.2023 21:40
$(n-1)^2+n^2+(n+1)^2=1111a$, where $1 \leq a \leq 9$ $(n-1)^2+n^2+(n+1)^2=3n^2+2 \equiv 2 \pmod{3}$ Now we check which multiples of 1111 have a remainder of $2 \pmod {3}$ $1111 \equiv 1 \pmod{3} \Rightarrow 2222,5555,8888 \equiv 2 \pmod{3}$ $3n^2+2=2222 \Rightarrow n^2=740 \Rightarrow n= \sqrt{740} \rightarrow \leftarrow$ $3n^2+2=5555 \Rightarrow n^2=1851 \Rightarrow n= \sqrt{1851} \rightarrow \leftarrow$ $3n^2+2=8888 \Rightarrow n^2=2962 \Rightarrow n= \sqrt{2962} \rightarrow \leftarrow$ Thus, there are no solutions $\blacksquare$