SAT Checking Cases

Correct Answer: D

Solution:

Tip: Recognize first few perfect squares $(1, 4, 9, ... 400)$ and cubes $(1, 8, 27,... 1000).$
Notice that if $|m| \geq 8$ , then $m^2 \geq 64$ and so $m^2 + n^2 \geq 64$ and thus there are no solutions.

If $m = \pm 7$ , then $n^2 = 50 - 49 = 1$ which has the solutions $n = 1, -1$ .
If $m = \pm 6$ , then $n^2 = 50 - 36 = 14$ which has no integer solutions.
If $m = \pm 5$ , then $n^2 = 50 - 25 = 25$ which has the solutions $n = 5, -5$ .
If $m = \pm 4$ , then $n^2 = 50 - 16 = 34$ which has no integer solutions.
If $m = \pm 3$ , then $n^2 = 50 - 9 = 41$ which has no integer solutions.
If $m = \pm 2$ , then $n^2 = 50 - 4 = 46$ which has no integer solutions.
If $m = \pm 1$ , then $n^2 = 50 - 1 = 49$ which has the solutions $n = 7, -7$ .
If $m = 0$ , then $n^2 = 50 - 0 = 50$ which has no integer solutions.

Hence, there are 12 integer solutions altogether (don't forget to count both the positive and negative case for $m$ ).

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Incorrect Choices:

(A) , (B) , (D) , and (E)
The solution explains how to eliminate these choices.

If you got this problem wrong, you should review SAT Numbers

The pairs of integers are (1,7),(7,1),(-1,-7),(-1,7),(7,-1),(-7,1),(-7,-1),(-7,1),(1,-7),(5,5),(-5,-5),(5,-5),(-5,5). Therefore the number of pairs of integers is 12.