Smallest positive integer
[vc_row][vc_column][vc_column_text]$1^2+2^2=5$, which is not a square. $1^2+2^2+3^2=14$, which is not a square. What is the smallest positive integer value of $n, n>1,$ such that $1^2+2^2+ \dots +n^2$ is a square number? Are there any larger possible values of $n$?[latexpage][/vc_column_text][lvca_accordion][lvca_panel]To show that $\sum\limits_{r=1}^{n}{{{r}^{2}}}={{N}^{2}}$ has a unique solution in integers for $n>1$.
Since $\operatorname{hcf}\left( m,n \right)=\operatorname{hcf}\left( m,m+n \right)$, with $m=n+1$ we have that $n,n+1$ and $2n+1$ are co-prime. For $\sum\limits_{r=1}^{n}{{{r}^{2}}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$ to be square the (co-prime) terms in the numerator must be square numbers multiplied by coefficients with product 6.| $n$ | $n+1$ | $2n+1$ | Reason why not possible |
|---|---|---|---|
| $6{{a}^{2}}$ | ${{b}^{2}}$ | ${{c}^{2}}$ | |
| ${{a}^{2}}$ | $6{{b}^{2}}$ | ${{c}^{2}}$ | $n+1=6{{b}^{2}}\Rightarrow 2n+1=12{{b}^{2}}-1$ which can't be square since congruent to 11(mod12). |
| ${{a}^{2}}$ | ${{b}^{2}}$ | $6{{c}^{2}}$ | $2n+1$ is an odd number but $6{{c}^{2}}$ is even. |
| ${{a}^{2}}$ | $2{{b}^{2}}$ | $3{{c}^{2}}$ | $n+1=2{{b}^{2}}\Rightarrow 2n+1=4{{b}^{2}}-1$ which can't be square since congruent to 3(mod4). |
| ${{a}^{2}}$ | $3{{b}^{2}}$ | $2{{c}^{2}}$ | $2n+1$ is an odd number but $2{{c}^{2}}$ is even. |
| $2{{a}^{2}}$ | ${{b}^{2}}$ | $3{{c}^{2}}$ | $n=2{{a}^{2}}\Rightarrow 2n+1=4{{a}^{2}}+1$ but this is not divisible by 3: $4{{\left( 3k\pm 1 \right)}^{2}}+1\equiv 2\left( \bmod 3 \right)$ and $4{{\left( 3k \right)}^{2}}+1\equiv 1\left( \bmod 3 \right)$. |
| $2{{a}^{2}}$ | $3{{b}^{2}}$ | ${{c}^{2}}$ | $n+1=3{{b}^{2}}\Rightarrow 2n+1=6{{b}^{2}}-1$ which can't be square since congruent to 5(mod6). |
| $3{{a}^{2}}$ | ${{b}^{2}}$ | $2{{c}^{2}}$ | $2n+1$ is an odd number but $2{{c}^{2}}$ is even. |
| $3{{a}^{2}}$ | $2{{b}^{2}}$ | ${{c}^{2}}$ | $n+1=2{{b}^{2}}\Rightarrow 2n+1=4{{b}^{2}}-1$ which can't be square since congruent to 3(mod4). |
| $c-b$ | $c+b$ | Reason why not possible |
|---|---|---|
| 1 | $6{{a}^{2}}$ | $2c=1+6{{a}^{2}}$ would not give an integer for $c$. |
| 2 | $3{{a}^{2}}$ | |
| 3 | $2{{a}^{2}}$ | $2c=3+2{{a}^{2}}$ would not give an integer for $c$. |
| 6 | ${{a}^{2}}$ | ${{c}^{2}}={{\left( b+6 \right)}^{2}}={{b}^{2}}+12b+36$. This leads to $n=12b+36$ and so ${{b}^{2}}-1=12b+36$ which does not have integer solutions. |
| ${{a}^{2}}$ | 6 | $c>b>1$ . Only possibility is $c=4,b=2$ but doesn't lead to a solution. |
