In a family there are four children of diļ¬erent ages, each age being a positive integer not less than $2$ and not greater than $16$. A year ago the square of the age of the eldest child was equal to the sum of the squares of the ages of the remaining children. One year from now the sum of the squares of the youngest and the oldest will be equal to the sum of the squares of the other two. How old is each child?
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Tags: inequalities, number theory unsolved, number theory
Solar Plexsus
27.12.2010 18:16
Let $a$, $b$, $c$ and $d$ be the ages in years of the four children where \begin{eqnarray} 1 < a < b < c < d < 17. \end{eqnarray} From the information given we get the following two equations: \begin{eqnarray} \addtocounter{equation}{1} (d - 1)^2 & = & (c - 1)^2 + (b - 1)^2 + (a - 1)^2 \\ (a + 1)^2 + (d+1)^2 & = & (c + 1)^2 + (b + 1)^2 \end{eqnarray} By calculating equation (3) - equation (2) and equation (3) + equation (2) respectively, the results are \begin{eqnarray} \addtocounter{equation}{3} a^2 + 1 & = & 2(b + c - d) \\ 2a - 1 & = & b^2 + c^2 - d^2 \end{eqnarray} From the ineqalities (1) we observe that $c \leq d - 1$ and $b \leq d - 2$. These two facts combined with equation (4) yields $a^2 + 1 \; = \; 2(b + c - d) \; \leq \; 2((d - 2) + (d - 1) - d) \; = \; 2(d - 3),$ hence $a^2 \; \leq \; 2d - 7 \; \leq \; 2 \cdot 16 - 7 = 32 - 7 = 25 = 5^2.$ Therefore $a \leq 5$ with equality (i.e. $a = 5$) when $d = 16$, $c = 15$ and $b = 14$ which is not possible since $(a,b,c,d) = (5,14,15,16)$ does not satisfies equation (5). Therefore $1 < a < 5$, which implies $a = 3$ because $a$ is odd by equation (4). Setting $a = 3$ in the equations (4) and (5) we get \begin{eqnarray} \addtocounter{equation}{5} b + c - d & = & 5 \\ b^2 + c^2 - d^2 & = & 5 \end{eqnarray} From equation (6) we have $d = b + c - 5$, which inserted in equation (7) give us \begin{eqnarray*} b^2 + c^2 - (b + c - 5)^2 & = & 5. \end{eqnarray*} This equation is equivalent to \begin{eqnarray} \addtocounter{equation}{7} (b - 5)(c - 5) & = & 10. \end{eqnarray} Now $c \geq a + 2 = 3 + 2 = 5$ by the inequalities (1), hence $c > b > 5$ by equation (8). Using equation (8) once again we see there are only two possibilities, namely $b - 5 = 1 \; \wedge \; c - 5 = 10,$ $b - 5 = 2 \; \wedge \; c - 5 = 5.$ Therefore $(b,c)=(6,15)$ or $(b,c)=(7,10)$, and recalling that $a=3$ and $d = b + c - 5$, we find that $(a,b,c,d)=(3,6,15,16)$ and $(a,b,c,d)=(3,7,10,12)$ are the only possible solutions of the equations (2)-(3) which satiesfies the inequalities (1). Hence this problem has two solutions: The four children are either 3, 6, 15 and 16 years old or they are 3, 7, 10 and 12 years old.