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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BobsonJoe
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"?
kvedula2004
22.06.2017 01:32
Let the numbers be $n-1,n,n+1$. The sum of their squares is $3n^2+2$. This must be congruent to $0 (mod 1111)$. Using the discriminant method to solve quadratic congruences, we see that all solution are far too large to make a four digit number (Lowest: $71^2+72^2+73^2=15554$) so there are no solutions. This is very rushed so please correct if needed. I will also post the method for solving quadratic congruences if needed as well.
FedeX333X
23.06.2017 02:22
Let the three numbers be $n-1, n, n+1$. The sum of their squares is $3n^2+2$, and we have $3n^2+2=1111k$ with $1 \le k \le 9$. We see that $3n^2+2 \equiv 2 \pmod{3}$, so also $1111k \equiv 2 \pmod{3}$, and this means that $k \equiv 2 \pmod{3}$, so $k=2,5,8$. We can then try solving it, or we could look at the equation with $\pmod{4}$: remember that a square is only $\equiv 0, 1 \pmod{4}$, hence the only "good" value of $k$ is $2$, but we can easily verify that $740$ is not a perfect square.
Rijadinho
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 $\blacksquare$