Suppose that $a, b, c, d$ and $e$ are consecutive positive integers with $a<b<c<d<e$. If $a^{2}+b^{2}+c^{2}=d^{2}+e^{2}$, what is the value of $a$ ?
Solution 1

Since $a, b, c, d$, and $e$ are consecutive with $a<b<c<d<e$, we can write $b=a+1$ and $c=a+2$ and $d=a+3$ and $e=a+4$.

From $a^{2}+b^{2}+c^{2}=d^{2}+e^{2}$, we obtain the equivalent equations:

$$
\begin{aligned}
a^{2}+(a+1)^{2}+(a+2)^{2} & =(a+3)^{2}+(a+4)^{2} \\
a^{2}+a^{2}+2 a+1+a^{2}+4 a+4 & =a^{2}+6 a+9+a^{2}+8 a+16 \\
a^{2}-8 a-20 & =0 \\
(a-10)(a+2) & =0
\end{aligned}
$$

Since $a$ is positive, then $a=10$.

(Checking, we see that $10^{2}+11^{2}+12^{2}=100+121+144=365$ and $13^{2}+14^{2}=169+196=365$.)

## Solution 2

Since $a, b, c, d$, and $e$ are consecutive with $a<b<c<d<e$, we can write $b=c-1$ and $a=c-2$ and $d=c+1$ and $e=c+2$.

From $a^{2}+b^{2}+c^{2}=d^{2}+e^{2}$, we obtain the equivalent equations:

$$
\begin{aligned}
(c-2)^{2}+(c-1)^{2}+c^{2} & =(c+1)^{2}+(c+2)^{2} \\
c^{2}-4 c+4+c^{2}-2 c+1+c^{2} & =c^{2}+2 c+1+c^{2}+4 c+4 \\
c^{2}-12 c & =0 \\
c(c-12) & =0
\end{aligned}
$$

Since $c$ is a positive integer, then $c=12$, which means that $a=c-2=10$.

(Checking, we see that $10^{2}+11^{2}+12^{2}=100+121+144=365$ and $13^{2}+14^{2}=169+196=365$.)