MHB Integer Solutions - Solve $a^2+b^2+c^2 + a + b+ c = 1$

[kaliprasad]

Solve for integers $a,b,c$ given $a^2+b^2+c^2 + a + b+ c = 1$

 

[anemone]

My solution:

Spoiler

Note that from the Cauchy-Schwarz inequality and the formula $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$, we have the following inequality that always holds for all real $a,\,b$ and $c$:

$$a^2+b^2+c^2\ge ab+bc+ca$$

$$a^2+b^2+c^2\ge \frac{(a+b+c)^2}{3}$$(*)

But from the given equality, we know $a^2+b^2+c^2=1-(a+b+c)$. Replacing this relation into the inequality (*) and solve it for $a+b+c$, we see that we get:

$$1-(a+b+c)\ge \frac{(a+b+c)^2}{3}$$

$$3-3(a+b+c)\ge (a+b+c)^2$$

$$0\ge (a+b+c)^2+3(a+b+c)-3$$

$-3\le a+b+c \le 0$

But observe that all of the following cases don't yield for integer solutions for the system:

$a+b+c=-3\cap a^2+b^2+c^2=4;\,a+b+c=-2\cap a^2+b^2+c^2=3$

$a+b+c=-1\cap a^2+b^2+c^2=2;\,a+b+c=0\cap a^2+b^2+c^2=1$

Therefore there are no such integer in $a,\,b$ and $c$ such that $a^2+b^2+c^2 + a + b+ c = 1$.

Thanks kaliprasad for posing challenge for us! I appreciate that!(Cool)

 

[lfdahl]

My solution:

Spoiler

\[a^2+b^2+c^2+a+b+c = \left ( a+\frac{1}{2} \right )^2+\left ( b+\frac{1}{2} \right )^2+\left ( c+\frac{1}{2} \right )^2-\frac{3}{4} = 1\]

\[\Rightarrow (2a+1)^2+(2b+1)^2+(2c+1)^2= 7\]

In order to show, that there are no possible integer solutions, both of the following arguments apply:I. The only perfect squares contained in $7$ are $1^2=1$ and $2^2=4$. Any triple combination (e.g. $1 + 1 + 4$) of them does not yield $7$. II. For any integer $n$: $(2n+1)^2 \in \left \{ 1,9,25,... \right \}$, leaving only one allowed perfect square, namely $1$, which yields the triple sum $3$.

 

[kaliprasad]

My solution:

Spoiler

\[a^2+b^2+c^2+a+b+c = \left ( a+\frac{1}{2} \right )^2+\left ( b+\frac{1}{2} \right )^2+\left ( c+\frac{1}{2} \right )^2-\frac{3}{4} = 1\]

\[\Rightarrow (2a+1)^2+(2b+1)^2+(2c+1)^2= 7\]

In order to show, that there are no possible integer solutions, both of the following arguments apply:I. The only perfect squares contained in $7$ are $1^2=1$ and $2^2=4$. Any triple combination (e.g. $1 + 1 + 4$) of them does not yield $7$. II. For any integer $n$: $(2n+1)^2 \in \left \{ 1,9,25,... \right \}$, leaving only one allowed perfect square, namely $1$, which yields the triple sum $3$.

above solution is good and so also by anemone

mine is same as above except

Spoiler

taking mod 8 we have LHS = 3 and RHS = 7 and hence no solution