Suppose that $a, b, c$, and $d$ are positive integers which are not necessarily distinct. If $a^{2}+b^{2}+c^{2}+d^{2}=70$, what is the largest possible value of $a+b+c+d ?$
Since $8^{2}=6470$, the largest that any of the integers can possibly be is 8 . The first 8 positive perfect squares are $1^{2}=1,2^{2}=4,3^{2}=9,4^{2}=16,5^{2}=25,6^{2}=36$, $7^{2}=49$, and $8^{2}=64$.

The table below records the sum of the squares of any two integers between 1 and 8 inclusive. That is, for example, the cell corresponding to the row for 9 and the column for 25 contains the number $5^{2}+3^{2}=25+9=34$.

|  | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 |
| :---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 1 | 2 | 5 | 10 | 17 | 26 | 37 | 50 | 65 |
| 4 |  | 8 | 13 | 20 | 29 | 40 | 53 | 68 |
| 9 |  |  | 18 | 25 | 34 | 45 | 58 | 73 |
| 16 |  |  |  | 32 | 41 | 52 | 65 | 80 |
| 25 |  |  |  |  | 50 | 61 | 74 | 89 |
| 36 |  |  |  |  |  | 72 | 85 | 100 |
| 49 |  |  |  |  |  |  | 98 | 113 |
| 64 |  |  |  |  |  |  |  | 128 |

We now note that the sum of the squares of any two of $a, b, c$, and $d$ must be at least 2 , so the sum of any two of $a, b, c$, and $d$ cannot by more than $70-2=68$. Therefore, the possible sums of squares of two of $a, b, c$, and $d$ are the sums from the above table that do not exceed 68 :

$$
2,5,8,10,13,17,18,20,25,26,29,32,34,37,40,41,45,50,52,53,58,61,65,68
$$

To find quadruples $(a, b, c, d)$ satisfying $a^{2}+b^{2}+c^{2}+d^{2}=70$, we can find pairs of numbers in the list above that sum to 70 and then use the table to find $a, b c$, and $d$.

The pairs that sum to 70 are 2 and 68,5 and 65, 17 and 53, 18 and 52, 20 and 50, 25 and 45, 29 and 41.

The numbers 2 and 68 come from $1^{2}+1^{2}$ and $2^{2}+8^{2}$, so we could have that the four integers $a, b, c$, and $d$ are $1,1,2$, and 8 .

The numbers 5 and 65 come from $1^{2}+2^{2}$ and either $1^{2}+8^{2}$ or $4^{2}+7^{2}$. This means $a, b, c$, and $d$ could be either $1,1,2$, and 8 or $1,2,4$, and 7 . We have already found the first of these possibilities.

Continuing in this way, we must have that $a, b, c$, and $d$ are one of

$$
1,1,2,8 ; 1,2,4,7 ; 3,3,4,6 ; 2,4,5,5
$$

and the sums of the entries in these lists are 12,14,16, and 16, respectively. Therefore, the largest possible sum is $a+b+c+d=16$ (and it is achievable in two different ways).

ANSWER: 16