<Question>
Find the number of ordered pairs $(a,b)$ of integers such that $|a + bi| \le 5.$
</Question>

<think>
<Plan_1>
Okay, so I need to find the number of ordered pairs (a, b) of integers where the absolute value (or modulus) of the complex number a + bi is less than or equal to 5. Hmm, let me think. The modulus of a complex number a + bi is given by the square root of (a² + b²). So, the condition |a + bi| ≤ 5 translates to sqrt(a² + b²) ≤ 5. If I square both sides to eliminate the square root, that gives me a² + b² ≤ 25. So essentially, I need to find all integer pairs (a, b) such that a² + b² is less than or equal to 25.

Alright, so the problem now reduces to counting all integer solutions (a, b) to the inequality a² + b² ≤ 25. Since a and b are integers, their squares will be non-negative integers, and the sum has to be at most 25. Let me visualize this. The set of all such (a, b) pairs would lie inside or on the boundary of a circle with radius 5 centered at the origin in the coordinate plane. Since we're dealing with integers, we need to count all the lattice points inside this circle.

Let me approach this systematically. Since the circle is symmetric in all four quadrants, maybe I can compute the number of points in one quadrant and then multiply by four, adjusting for points on the axes which are shared between quadrants. But perhaps a simpler way is to fix each possible value of a and then determine the possible integer values of b such that a² + b² ≤ 25.

<prompt_1.1> Plan how to systematically count the integer solutions to a² + b² ≤ 25. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's consider that a and b can be positive or negative, but their squares make them non-negative. So, for each a from -5 to 5, I can find the range of b such that when squared and added to a², the sum is ≤25.

Wait, but since squaring removes the sign, maybe it's easier to consider all possible non-negative a and b, and then account for the signs accordingly. Let me think. For example, if a is positive, then the negative values of a would mirror the positive ones, except when a is zero. Similarly for b. So perhaps I can calculate the number of solutions in the first quadrant (where a and b are non-negative) and then multiply by 4 to account for all four quadrants, then adjust for the axes which are counted in two quadrants each, and finally add the origin which is only one point.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But maybe that's complicating it. Alternatively, since a and b can be positive or negative, for each a from -5 to 5, I can compute the possible b's. Let me try that approach.

First, let's note that a² must be ≤25, so a can range from -5 to 5 inclusive. Similarly for b. So, for each integer a from -5 to 5, I can compute the maximum b such that b² ≤25 - a². Then, for each a, the possible b's are from -floor(sqrt(25 - a²)) to floor(sqrt(25 - a²)), inclusive. Then, the number of b's for each a is 2*floor(sqrt(25 - a²)) +1 (to account for zero and the positive and negative values).

<prompt_2.1> Execute the plan to count the number of integer solutions for each a from -5 to 5. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Wait, actually, if the maximum |b| is k, then the number of integer b's is 2k +1 (from -k to k inclusive). So yes, if sqrt(25 - a²) is, say, k, then b can be from -k to k, which is 2k +1 numbers. But sqrt(25 - a²) might not be an integer. Wait, but 25 - a² is an integer, so sqrt(25 - a²) is either integer or irrational. Since we need b to be integer, so the maximum integer b is floor(sqrt(25 - a²)). Therefore, for each a, the number of b's is 2*floor(sqrt(25 - a²)) +1.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But let's check with an example. Let's take a =0. Then 25 -0 =25, so sqrt(25)=5. So floor(sqrt(25))=5. Then number of b's is 2*5 +1=11. Which makes sense: b can be -5, -4,....,0,....,4,5, total 11 numbers. Similarly, if a=3, then 25 -9=16, sqrt(16)=4, so floor is 4, so number of b's is 2*4 +1=9. Which would be b from -4 to 4.

So, the plan is:

1. For each integer a from -5 to 5:

<prompt_3.1> Follow the above plan to calculate the number of ordered pairs (a, b). </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
2. Compute k = floor(sqrt(25 - a²))

3. The number of b's for this a is 2k +1

4. Sum over all a.

However, since a is squared, positive and negative a's will give the same k. For example, a=1 and a=-1 both give 25 -1=24, sqrt(24)≈4.898, floor is 4, so number of b's is 2*4 +1=9 for each. Similarly, a=2 and a=-2: 25 -4=21, sqrt(21)≈4.583, floor 4, same as above. a=3 and -3: 25 -9=16, sqrt=4, so 9. a=4 and -4:25 -16=9, sqrt=3, floor=3, so 2*3 +1=7. a=5 and -5:25 -25=0, sqrt=0, so 1.

Therefore, instead of iterating through all a from -5 to 5, we can note that for each positive a, there is a corresponding negative a with the same number of b's, except when a=0. So maybe compute for a=0, then a=1, a=2, a=3, a=4, a=5, and multiply the counts for positive a's by 2 (to account for negative a's), then add the count for a=0.

Wait, but when a=0, it's only one point, not two. So let's structure it as:

Total pairs = (number of pairs for a=0) + 2*(number of pairs for a=1 + a=2 + a=3 + a=4 + a=5)

But let's compute each a:

First, a=0:

a=0: 25 -0=25, sqrt(25)=5, k=5, so number of b's=2*5 +1=11. So 11 pairs.

a=1:

a=1: 25 -1=24, sqrt(24)≈4.898, floor=4, so number of b's=2*4 +1=9. Similarly, a=-1 also gives 9. So total for a=1 and a=-1: 2*9=18.

Similarly, a=2:

a=2: 25 -4=21, sqrt(21)≈4.583, floor=4, so number of b's=9. So a=2 and a=-2: 2*9=18.

a=3:

a=3: 25 -9=16, sqrt=4, so number of b's=9. So a=3 and a=-3: 2*9=18.

a=4:

a=4:25 -16=9, sqrt=3, so number of b's=2*3 +1=7. So a=4 and a=-4:2*7=14.

a=5:

a=5:25 -25=0, sqrt=0, so number of b's=1. So a=5 and a=-5:2*1=2.

Wait, but hold on: when a=5, b can only be 0. Similarly for a=-5, b=0. So each of these is one point, so two points total.

Therefore, adding all these up:

a=0:11

a=±1:18

a=±2:18

a=±3:18

a=±4:14

a=±5:2

Total=11 +18 +18 +18 +14 +2= Let's compute step by step:

Start with 11.

11 +18=29

29 +18=47

47 +18=65

65 +14=79

79 +2=81.

So total number of ordered pairs is 81.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Wait, but let me verify this, because sometimes when you do these symmetric counts, it's easy to make a mistake. Let me check each a again:

<prompt_4.1> Verify the calculation of the total number of ordered pairs. </prompt_4.1>

But let me cross-verify this with another approach. Let's consider that the number of integer solutions to a² + b² ≤25 is the same as the number of lattice points inside a circle of radius 5. There is a formula called Gauss's circle problem which approximates the number of lattice points as πr² + error term, but for r=5, we can compute exactly.

Alternatively, another way to compute is to iterate over all possible a from -5 to 5 and for each a, compute the number of b's as we did before.

<prompt_4.2> Cross-verify the result using an alternative method. </prompt_4.2>

But let me check with another method. For example, in the first quadrant (a ≥0, b ≥0), the number of points is the number of (a, b) with a² + b² ≤25. Then, multiplying by 4 gives the points in all quadrants, excluding the axes which are counted in each quadrant. Wait, actually, the axes points are shared. For example, the point (a, 0) is on the x-axis and is part of the first and fourth quadrants. Similarly, (0, b) is on the y-axis, part of the first and second quadrants. The origin (0,0) is only one point.

Therefore, perhaps a better approach is:

Compute the number of points in the first quadrant (a>0, b>0), then multiply by 4.

Compute the number of points on the x-axis (a>0, b=0), multiply by 2 (since they are also on the fourth quadrant).

Compute the number of points on the y-axis (a=0, b>0), multiply by 2 (since they are also on the second quadrant).

Add the origin.

But let's see:

First, compute the number of points in the first quadrant (a>0, b>0) with a² + b² ≤25.

For a=1 to 5:

a=1: b² ≤24, so b can be 1 to 4 (since 4²=16 ≤24, 5²=25>24). So b=1,2,3,4: 4 points.

a=2: b² ≤21, so b=1 to 4 (same as above): 4 points.

a=3: b² ≤16, so b=1 to 4 (since 4²=16 ≤16): 4 points.

a=4: b² ≤9, so b=1 to 3 (since 3²=9 ≤9): 3 points.

a=5: b² ≤0, so no points (since b>0). So total in first quadrant: 4+4+4+3=15.

Multiply by 4: 15*4=60.

Then, points on the x-axis (a>0, b=0): a=1 to5: 5 points. Multiply by 2 (for a>0 and a<0): 5*2=10.

Points on the y-axis (a=0, b>0): b=1 to5:5 points. Multiply by 2 (for b>0 and b<0):5*2=10.

Add the origin:1.

Total:60 +10 +10 +1=81. Same result. So this confirms the total is 81.

Therefore, the answer should be 81.

But just to be thorough, let me list out all possible a from -5 to5 and count the b's manually.

<prompt_4.3> Manually list and count the ordered pairs to confirm the result. </prompt_4.3>
</Plan_4>

<Execution_4>
<execution_4.1>
For a=0: 11. Correct, since b can be -5 to 5.

For a=1: 2*4 +1=9. So b can be from -4 to 4. That's 9 values. Similarly for a=-1.

Same for a=2: 2*4 +1=9. Correct, since 25 -4=21, sqrt(21) is ~4.58, so floor 4. So b from -4 to 4.

a=3: 25 -9=16, sqrt=4. So b from -4 to 4. 9 values.

a=4:25 -16=9, sqrt=3. So b from -3 to 3. 7 values.

a=5:25 -25=0, sqrt=0. So b=0. 1 value.

Yes, that seems correct.

So adding all these up:

a=0:11

a=±1:9+9=18

a=±2:9+9=18

a=±3:9+9=18

a=±4:7+7=14

a=±5:1+1=2

Total:11 +18 +18 +18 +14 +2=81.
</execution_4.1>

<execution_4.2>
Alternatively, since a and b can be positive or negative, but their squares are the same, we can compute the number of solutions in the first quadrant and multiply accordingly, but as we saw earlier, the first method of iterating through each a and counting the b's gives 81. Since both methods give the same result, that gives me more confidence.

Alternatively, let's think of it as a square grid from (-5,-5) to (5,5). For each a from -5 to5, the number of b's is 2*floor(sqrt(25 -a²)) +1. So let's compute this for each a:

For a= -5: same as a=5. a²=25. sqrt(25 -25)=0. floor(0)=0. So 2*0 +1=1. So b=0. So ( -5,0). Similarly for a=5: (5,0). So two points.

For a= -4: a²=16. 25 -16=9. sqrt(9)=3. floor(3)=3. So 2*3 +1=7. So b can be -3,-2,-1,0,1,2,3. So 7 points. Similarly for a=4: same.

For a= -3: a²=9. 25 -9=16. sqrt(16)=4. floor(4)=4. So 2*4 +1=9. So b from -4 to4. Similarly for a=3.

For a= -2: a²=4. 25 -4=21. sqrt(21)=4.583, floor=4. So 2*4 +1=9. So same as above.

For a= -1: a²=1. 25 -1=24. sqrt(24)=4.898, floor=4. So 2*4 +1=9.

For a=0: 25 -0=25. sqrt=5. floor=5. So 2*5 +1=11.

Therefore, for each a:

a=-5:1

a=-4:7

a=-3:9

a=-2:9

a=-1:9

a=0:11

a=1:9

a=2:9

a=3:9

a=4:7

a=5:1

Adding all these up:

1 +7=8

8 +9=17

17 +9=26

26 +9=35

35 +11=46

46 +9=55

55 +9=64

64 +9=73

73 +7=80

80 +1=81. Yes, same total. So that's correct.

Therefore, the number of ordered pairs is 81.
</execution_4.2>

<execution_4.3>
Starting with a=0:

a=0: b² ≤25 => b from -5 to5: 11 points.

a=1:

a²=1, so b² ≤24. The maximum b is 4 (since 4²=16 ≤24, 5²=25>24). So b can be -4,-3,-2,-1,0,1,2,3,4: 9 points.

Similarly for a=-1: same number of b's:9.

a=2:

a²=4, so b² ≤21. sqrt(21)≈4.58, so b can be -4 to4:9 points.

a=-2: same:9.

a=3:

a²=9, so b² ≤16. sqrt(16)=4. So b can be -4 to4:9 points.

a=-3: same:9.

a=4:

a²=16, so b² ≤9. sqrt(9)=3. So b can be -3,-2,-1,0,1,2,3:7 points.

a=-4: same:7.

a=5:

a²=25, so b² ≤0. So b=0. 1 point.

a=-5: same:1.

So adding up:

a=0:11

a=±1:9+9=18

a=±2:9+9=18

a=±3:9+9=18

a=±4:7+7=14

a=±5:1+1=2

Total:11 +18 +18 +18 +14 +2=81. Same result. So that's consistent.

Therefore, the number of ordered pairs (a, b) of integers such that |a + bi| ≤5 is 81.
</execution_4.3>
</Execution_4>

<Final_answer>
Therefore, the number of ordered pairs of integers $(a, b)$ such that $|a + bi| \le 5$ is $\boxed{81}$.
</Final_answer>
</think>