- #1
kaliprasad
Gold Member
MHB
- 1,335
- 0
Solve for integers $a,b,c$ given $a^2+b^2+c^2 + a + b+ c = 1$
lfdahl said:My solution:
\[a^2+b^2+c^2+a+b+c = \left ( a+\frac{1}{2} \right )^2+\left ( b+\frac{1}{2} \right )^2+\left ( c+\frac{1}{2} \right )^2-\frac{3}{4} = 1\]
\[\Rightarrow (2a+1)^2+(2b+1)^2+(2c+1)^2= 7\]
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 $3$.
Integer solutions are values of the variables (a, b, and c) that satisfy the given equation and are also integers, which are whole numbers (positive, negative, or zero) without any fractions or decimals.
To solve for integer solutions, we can use techniques such as substitution, elimination, or graphing to find the values of the variables that make the given equation true. In this particular equation, we can use the fact that all the terms are squared to our advantage and try different combinations of integers until we find a solution.
Unfortunately, there is no guaranteed shortcut to finding integer solutions. However, some equations may have patterns or special properties that can make finding solutions easier. It is always best to try different techniques and methods to find solutions.
Yes, there can be multiple integer solutions to this equation. In fact, there can be infinite solutions, as long as the values of a, b, and c satisfy the equation and are also integers. This is because there are an infinite number of integers that can be squared and added together to equal 1.
Yes, negative integers can be used as solutions in this equation. In fact, some equations may only have negative integer solutions. It is important to consider all possible values of the variables when looking for integer solutions.