Prove that given any five integers, there will be three for which the sum of the squares of those integers is divisible by 3.

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Basically the question is asking us to prove that given any integers $$x_1,x_2,x_3,x_4,x_5$$ Prove that 3 of the integers from the set above, suppose $$x_a,x_b,x_c$$ satisfy this equation: $$x_a^2 + x_b^2 + x_c^2 = 3k$$ So I know I am suppose to use the pigeon hole principle to prove this. I know that if I have 5 pigeons and 2 holes then 1 hole will have 3 pigeons. But what I am confused about is how do you define the hole? Do I just say that the container has a property such that if 3 integers are in it then those 3 integers squared sum up to a multiple of 3?










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    Basically the question is asking us to prove that given any integers $$x_1,x_2,x_3,x_4,x_5$$ Prove that 3 of the integers from the set above, suppose $$x_a,x_b,x_c$$ satisfy this equation: $$x_a^2 + x_b^2 + x_c^2 = 3k$$ So I know I am suppose to use the pigeon hole principle to prove this. I know that if I have 5 pigeons and 2 holes then 1 hole will have 3 pigeons. But what I am confused about is how do you define the hole? Do I just say that the container has a property such that if 3 integers are in it then those 3 integers squared sum up to a multiple of 3?










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      Basically the question is asking us to prove that given any integers $$x_1,x_2,x_3,x_4,x_5$$ Prove that 3 of the integers from the set above, suppose $$x_a,x_b,x_c$$ satisfy this equation: $$x_a^2 + x_b^2 + x_c^2 = 3k$$ So I know I am suppose to use the pigeon hole principle to prove this. I know that if I have 5 pigeons and 2 holes then 1 hole will have 3 pigeons. But what I am confused about is how do you define the hole? Do I just say that the container has a property such that if 3 integers are in it then those 3 integers squared sum up to a multiple of 3?










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      Basically the question is asking us to prove that given any integers $$x_1,x_2,x_3,x_4,x_5$$ Prove that 3 of the integers from the set above, suppose $$x_a,x_b,x_c$$ satisfy this equation: $$x_a^2 + x_b^2 + x_c^2 = 3k$$ So I know I am suppose to use the pigeon hole principle to prove this. I know that if I have 5 pigeons and 2 holes then 1 hole will have 3 pigeons. But what I am confused about is how do you define the hole? Do I just say that the container has a property such that if 3 integers are in it then those 3 integers squared sum up to a multiple of 3?







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      asked 31 mins ago









      Geralt

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          Any square integer must be congruent to either 1 or 0 mod 3. So for each of the 5 squares, we put it into hole 0 if it is congruent to 0 and into hole 1 if it is congruent to 1. Then take three squares from the hole with at least 3 squares and add them together. You will get either: $0+0+0equiv 0$ or $1+1+1equiv0$ mod 3.






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            Let's look at any $3$ of them,say $a,b,c$ among $a,b,c,d,e$. You must have $2$ cases: $a^2 = 0, b^2=1, c^2=0$ or $a^2=0, b^2=1,c^2=1$ for the worst scenario. For the last $2$ numbers $d,e$, if at least one, say $d^2 = 0$, you're done. If not $d^2= e^2 = 1$, then $b^2+d^2+e^2 = 0$, all mod $3$. And you are done.






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              The remainder of every integer in dividing by $3$ is either $0$,$1$,or $2$.



              Thus the remainder of a square in dividing by $3$ is either $0$ or $1$.



              Now we have $5$ perfect squares that means a set of $5$ remainders each of which is either a $0$ or a $1$.



              The Pigeon hole principle says there are at least three of the same kind in the set of remainders.
              Well, we either have $1+1+1$ or $0+0+0$ in our sum of the squares and in either case the sum is divisible by $3$






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                down vote













                Basic pigeon hole. If all integers fall into either type $A$ or type $B$ and you have $n$ integers. Then at least $lceil frac n2 rceil$ will be of the same type.



                I hope you can convince yourself of that on your own.



                ...



                Let $x_i = 3*M_i + r_i$ where $r_i = 0, 1,$ or $-1$. All numbers will be one of these three options.



                Then $x_i^2 = 9*M_i^2 + 6*M_i*r_i + r_i^2$. Let $V_i = 3*M_i^2 + 2*M_i*r_i$ so $x_i^2 = 3V_i + r_i^2$ where $r_i^2 = 0$ or $1$. All numbers will be one of these $2$ options.



                Let all integers, $x_i$ where $r_i^2 = 0$ by of type $A$ and let all integers, $x_j$ where $r_j^2 =1$ by of type $B$.



                So if you have $5$ integers and each is of type $A$ or type $B$ then by pigeon hole you must have at least $3$ of same type.



                So suppose $x_a, x_b, x_c$ are three that are of one of these two types.



                If the three are of the type where $A$ where $r_i^2 = 0$ then you will $x_a^2 + x_b^2 + x_c^2 = 3V_a + 0 + 3V_b + 0 + 3V_c + 0$. And that is divisible by $3$.



                If the three are of the type $B$ where $r_i^2 = 1$ then you will have $x_a^2 + x_b^2 + x_c^2 = 3V_a + 1 + 3V_b+1 + 3V_c + 1 = 3V_a + 3V_b + 3V_c + 3$ which is divisible by $3$.



                So you will always have at least three integers whose sums of squares is divisible by $3$.






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                  Any square integer must be congruent to either 1 or 0 mod 3. So for each of the 5 squares, we put it into hole 0 if it is congruent to 0 and into hole 1 if it is congruent to 1. Then take three squares from the hole with at least 3 squares and add them together. You will get either: $0+0+0equiv 0$ or $1+1+1equiv0$ mod 3.






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                    up vote
                    5
                    down vote













                    Any square integer must be congruent to either 1 or 0 mod 3. So for each of the 5 squares, we put it into hole 0 if it is congruent to 0 and into hole 1 if it is congruent to 1. Then take three squares from the hole with at least 3 squares and add them together. You will get either: $0+0+0equiv 0$ or $1+1+1equiv0$ mod 3.






                    share|cite|improve this answer






















                      up vote
                      5
                      down vote










                      up vote
                      5
                      down vote









                      Any square integer must be congruent to either 1 or 0 mod 3. So for each of the 5 squares, we put it into hole 0 if it is congruent to 0 and into hole 1 if it is congruent to 1. Then take three squares from the hole with at least 3 squares and add them together. You will get either: $0+0+0equiv 0$ or $1+1+1equiv0$ mod 3.






                      share|cite|improve this answer












                      Any square integer must be congruent to either 1 or 0 mod 3. So for each of the 5 squares, we put it into hole 0 if it is congruent to 0 and into hole 1 if it is congruent to 1. Then take three squares from the hole with at least 3 squares and add them together. You will get either: $0+0+0equiv 0$ or $1+1+1equiv0$ mod 3.







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                      answered 24 mins ago









                      Ricky Tensor

                      2714




                      2714




















                          up vote
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                          Let's look at any $3$ of them,say $a,b,c$ among $a,b,c,d,e$. You must have $2$ cases: $a^2 = 0, b^2=1, c^2=0$ or $a^2=0, b^2=1,c^2=1$ for the worst scenario. For the last $2$ numbers $d,e$, if at least one, say $d^2 = 0$, you're done. If not $d^2= e^2 = 1$, then $b^2+d^2+e^2 = 0$, all mod $3$. And you are done.






                          share|cite|improve this answer
























                            up vote
                            1
                            down vote













                            Let's look at any $3$ of them,say $a,b,c$ among $a,b,c,d,e$. You must have $2$ cases: $a^2 = 0, b^2=1, c^2=0$ or $a^2=0, b^2=1,c^2=1$ for the worst scenario. For the last $2$ numbers $d,e$, if at least one, say $d^2 = 0$, you're done. If not $d^2= e^2 = 1$, then $b^2+d^2+e^2 = 0$, all mod $3$. And you are done.






                            share|cite|improve this answer






















                              up vote
                              1
                              down vote










                              up vote
                              1
                              down vote









                              Let's look at any $3$ of them,say $a,b,c$ among $a,b,c,d,e$. You must have $2$ cases: $a^2 = 0, b^2=1, c^2=0$ or $a^2=0, b^2=1,c^2=1$ for the worst scenario. For the last $2$ numbers $d,e$, if at least one, say $d^2 = 0$, you're done. If not $d^2= e^2 = 1$, then $b^2+d^2+e^2 = 0$, all mod $3$. And you are done.






                              share|cite|improve this answer












                              Let's look at any $3$ of them,say $a,b,c$ among $a,b,c,d,e$. You must have $2$ cases: $a^2 = 0, b^2=1, c^2=0$ or $a^2=0, b^2=1,c^2=1$ for the worst scenario. For the last $2$ numbers $d,e$, if at least one, say $d^2 = 0$, you're done. If not $d^2= e^2 = 1$, then $b^2+d^2+e^2 = 0$, all mod $3$. And you are done.







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                              answered 22 mins ago









                              DeepSea

                              69.7k54386




                              69.7k54386




















                                  up vote
                                  0
                                  down vote













                                  The remainder of every integer in dividing by $3$ is either $0$,$1$,or $2$.



                                  Thus the remainder of a square in dividing by $3$ is either $0$ or $1$.



                                  Now we have $5$ perfect squares that means a set of $5$ remainders each of which is either a $0$ or a $1$.



                                  The Pigeon hole principle says there are at least three of the same kind in the set of remainders.
                                  Well, we either have $1+1+1$ or $0+0+0$ in our sum of the squares and in either case the sum is divisible by $3$






                                  share|cite
























                                    up vote
                                    0
                                    down vote













                                    The remainder of every integer in dividing by $3$ is either $0$,$1$,or $2$.



                                    Thus the remainder of a square in dividing by $3$ is either $0$ or $1$.



                                    Now we have $5$ perfect squares that means a set of $5$ remainders each of which is either a $0$ or a $1$.



                                    The Pigeon hole principle says there are at least three of the same kind in the set of remainders.
                                    Well, we either have $1+1+1$ or $0+0+0$ in our sum of the squares and in either case the sum is divisible by $3$






                                    share|cite






















                                      up vote
                                      0
                                      down vote










                                      up vote
                                      0
                                      down vote









                                      The remainder of every integer in dividing by $3$ is either $0$,$1$,or $2$.



                                      Thus the remainder of a square in dividing by $3$ is either $0$ or $1$.



                                      Now we have $5$ perfect squares that means a set of $5$ remainders each of which is either a $0$ or a $1$.



                                      The Pigeon hole principle says there are at least three of the same kind in the set of remainders.
                                      Well, we either have $1+1+1$ or $0+0+0$ in our sum of the squares and in either case the sum is divisible by $3$






                                      share|cite












                                      The remainder of every integer in dividing by $3$ is either $0$,$1$,or $2$.



                                      Thus the remainder of a square in dividing by $3$ is either $0$ or $1$.



                                      Now we have $5$ perfect squares that means a set of $5$ remainders each of which is either a $0$ or a $1$.



                                      The Pigeon hole principle says there are at least three of the same kind in the set of remainders.
                                      Well, we either have $1+1+1$ or $0+0+0$ in our sum of the squares and in either case the sum is divisible by $3$







                                      share|cite












                                      share|cite



                                      share|cite










                                      answered 9 mins ago









                                      Mohammad Riazi-Kermani

                                      35.6k41955




                                      35.6k41955




















                                          up vote
                                          0
                                          down vote













                                          Basic pigeon hole. If all integers fall into either type $A$ or type $B$ and you have $n$ integers. Then at least $lceil frac n2 rceil$ will be of the same type.



                                          I hope you can convince yourself of that on your own.



                                          ...



                                          Let $x_i = 3*M_i + r_i$ where $r_i = 0, 1,$ or $-1$. All numbers will be one of these three options.



                                          Then $x_i^2 = 9*M_i^2 + 6*M_i*r_i + r_i^2$. Let $V_i = 3*M_i^2 + 2*M_i*r_i$ so $x_i^2 = 3V_i + r_i^2$ where $r_i^2 = 0$ or $1$. All numbers will be one of these $2$ options.



                                          Let all integers, $x_i$ where $r_i^2 = 0$ by of type $A$ and let all integers, $x_j$ where $r_j^2 =1$ by of type $B$.



                                          So if you have $5$ integers and each is of type $A$ or type $B$ then by pigeon hole you must have at least $3$ of same type.



                                          So suppose $x_a, x_b, x_c$ are three that are of one of these two types.



                                          If the three are of the type where $A$ where $r_i^2 = 0$ then you will $x_a^2 + x_b^2 + x_c^2 = 3V_a + 0 + 3V_b + 0 + 3V_c + 0$. And that is divisible by $3$.



                                          If the three are of the type $B$ where $r_i^2 = 1$ then you will have $x_a^2 + x_b^2 + x_c^2 = 3V_a + 1 + 3V_b+1 + 3V_c + 1 = 3V_a + 3V_b + 3V_c + 3$ which is divisible by $3$.



                                          So you will always have at least three integers whose sums of squares is divisible by $3$.






                                          share|cite|improve this answer


























                                            up vote
                                            0
                                            down vote













                                            Basic pigeon hole. If all integers fall into either type $A$ or type $B$ and you have $n$ integers. Then at least $lceil frac n2 rceil$ will be of the same type.



                                            I hope you can convince yourself of that on your own.



                                            ...



                                            Let $x_i = 3*M_i + r_i$ where $r_i = 0, 1,$ or $-1$. All numbers will be one of these three options.



                                            Then $x_i^2 = 9*M_i^2 + 6*M_i*r_i + r_i^2$. Let $V_i = 3*M_i^2 + 2*M_i*r_i$ so $x_i^2 = 3V_i + r_i^2$ where $r_i^2 = 0$ or $1$. All numbers will be one of these $2$ options.



                                            Let all integers, $x_i$ where $r_i^2 = 0$ by of type $A$ and let all integers, $x_j$ where $r_j^2 =1$ by of type $B$.



                                            So if you have $5$ integers and each is of type $A$ or type $B$ then by pigeon hole you must have at least $3$ of same type.



                                            So suppose $x_a, x_b, x_c$ are three that are of one of these two types.



                                            If the three are of the type where $A$ where $r_i^2 = 0$ then you will $x_a^2 + x_b^2 + x_c^2 = 3V_a + 0 + 3V_b + 0 + 3V_c + 0$. And that is divisible by $3$.



                                            If the three are of the type $B$ where $r_i^2 = 1$ then you will have $x_a^2 + x_b^2 + x_c^2 = 3V_a + 1 + 3V_b+1 + 3V_c + 1 = 3V_a + 3V_b + 3V_c + 3$ which is divisible by $3$.



                                            So you will always have at least three integers whose sums of squares is divisible by $3$.






                                            share|cite|improve this answer
























                                              up vote
                                              0
                                              down vote










                                              up vote
                                              0
                                              down vote









                                              Basic pigeon hole. If all integers fall into either type $A$ or type $B$ and you have $n$ integers. Then at least $lceil frac n2 rceil$ will be of the same type.



                                              I hope you can convince yourself of that on your own.



                                              ...



                                              Let $x_i = 3*M_i + r_i$ where $r_i = 0, 1,$ or $-1$. All numbers will be one of these three options.



                                              Then $x_i^2 = 9*M_i^2 + 6*M_i*r_i + r_i^2$. Let $V_i = 3*M_i^2 + 2*M_i*r_i$ so $x_i^2 = 3V_i + r_i^2$ where $r_i^2 = 0$ or $1$. All numbers will be one of these $2$ options.



                                              Let all integers, $x_i$ where $r_i^2 = 0$ by of type $A$ and let all integers, $x_j$ where $r_j^2 =1$ by of type $B$.



                                              So if you have $5$ integers and each is of type $A$ or type $B$ then by pigeon hole you must have at least $3$ of same type.



                                              So suppose $x_a, x_b, x_c$ are three that are of one of these two types.



                                              If the three are of the type where $A$ where $r_i^2 = 0$ then you will $x_a^2 + x_b^2 + x_c^2 = 3V_a + 0 + 3V_b + 0 + 3V_c + 0$. And that is divisible by $3$.



                                              If the three are of the type $B$ where $r_i^2 = 1$ then you will have $x_a^2 + x_b^2 + x_c^2 = 3V_a + 1 + 3V_b+1 + 3V_c + 1 = 3V_a + 3V_b + 3V_c + 3$ which is divisible by $3$.



                                              So you will always have at least three integers whose sums of squares is divisible by $3$.






                                              share|cite|improve this answer














                                              Basic pigeon hole. If all integers fall into either type $A$ or type $B$ and you have $n$ integers. Then at least $lceil frac n2 rceil$ will be of the same type.



                                              I hope you can convince yourself of that on your own.



                                              ...



                                              Let $x_i = 3*M_i + r_i$ where $r_i = 0, 1,$ or $-1$. All numbers will be one of these three options.



                                              Then $x_i^2 = 9*M_i^2 + 6*M_i*r_i + r_i^2$. Let $V_i = 3*M_i^2 + 2*M_i*r_i$ so $x_i^2 = 3V_i + r_i^2$ where $r_i^2 = 0$ or $1$. All numbers will be one of these $2$ options.



                                              Let all integers, $x_i$ where $r_i^2 = 0$ by of type $A$ and let all integers, $x_j$ where $r_j^2 =1$ by of type $B$.



                                              So if you have $5$ integers and each is of type $A$ or type $B$ then by pigeon hole you must have at least $3$ of same type.



                                              So suppose $x_a, x_b, x_c$ are three that are of one of these two types.



                                              If the three are of the type where $A$ where $r_i^2 = 0$ then you will $x_a^2 + x_b^2 + x_c^2 = 3V_a + 0 + 3V_b + 0 + 3V_c + 0$. And that is divisible by $3$.



                                              If the three are of the type $B$ where $r_i^2 = 1$ then you will have $x_a^2 + x_b^2 + x_c^2 = 3V_a + 1 + 3V_b+1 + 3V_c + 1 = 3V_a + 3V_b + 3V_c + 3$ which is divisible by $3$.



                                              So you will always have at least three integers whose sums of squares is divisible by $3$.







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                                              fleablood

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