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If \(a\) and \(b\) are integers, for how many ordered pairs \((a, b)\) is \(945 = a^2 - b^2\)?
A. 8
B. 16
C. 24
D. 32
E. 64
A. 8
B. 16
C. 24
D. 32
E. 64
07 Aug 2026, 05:06
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Official Solution:
If \(a\) and \(b\) are integers, for how many ordered pairs \((a, b)\) is \(945 = a^2 - b^2\)?
A. 8
B. 16
C. 24
D. 32
E. 64
\(945 = a^2 - b^2 = (a - b)(a + b)\)
Since 945 is positive, \(a - b\) and \(a + b\) must be either both positive or both negative.
Now factor 945:
\(945 = 3^3 * 5 * 7\)
So the number of positive divisors of 945 is:
\((3 + 1) * (1 + 1) * (1 + 1) = 16\)
For each positive divisor \(d\) of 945, we can set \(a - b = d\) and \(a + b = \frac{945}{d}\). Thus, there are 16 such ordered pairs with positive \(a - b\) and \(a + b\).
For example:
If \(a - b = 1\), then \(a + b = \frac{945}{1} = 945\).
Adding the two equations:
\(2a = 946\)
\(a = 473\)
Then \(b = 472\).
So this gives one ordered pair:
\((a, b) = (473, 472)\)
If \(a - b = 3\), then \(a + b = 315\).
Adding the two equations:
\(2a = 318\)
\(a = 159\)
Then \(b = 156\).
So this gives another ordered pair:
\((a, b) = (159, 156)\)
Also, reversing the factor values gives a different ordered pair.
For example:
If \(a - b = 945\), then \(a + b = 1\).
Adding the two equations:
\(2a = 946\)
\(a = 473\)
Then \(b = -472\).
So this gives:
\((a, b) = (473, -472)\)
Since 945 is odd, both \(d\) and \(\frac{945}{d}\) are odd, so the resulting values of \(a\) and \(b\) from each such pair will be integers. Thus, each positive divisor gives one ordered pair \((a, b)\). Since there are 16 positive divisors, we get 16 ordered pairs from positive factor pairs.
Similarly, the negatives \((-d, -\frac{945}{d})\) give another 16 ordered pairs of values for \(a - b\) and \(a + b\), with both values negative.
For example:
If \(a - b = -1\), then \(a + b = -945\).
Adding the two equations:
\(2a = -946\)
\(a = -473\)
Then \(b = -472\).
So the negative factor pairs give another 16 ordered pairs.
Therefore, the total number of ordered pairs \((a, b)\) is:
\(16 + 16 = 32\)
P.S. For reference, the 32 ordered pairs are:
\((473, 472), (159, 156), (97, 92), (71, 64), (57, 48), (39, 24), (33, 12), (31, 4)\)
\((31, -4), (33, -12), (39, -24), (57, -48), (71, -64), (97, -92), (159, -156), (473, -472)\)
\((-473, -472), (-159, -156), (-97, -92), (-71, -64), (-57, -48), (-39, -24), (-33, -12), (-31, -4)\)
\((-31, 4), (-33, 12), (-39, 24), (-57, 48), (-71, 64), (-97, 92), (-159, 156), (-473, 472)\)
Answer: D
If \(a\) and \(b\) are integers, for how many ordered pairs \((a, b)\) is \(945 = a^2 - b^2\)?
A. 8
B. 16
C. 24
D. 32
E. 64
\(945 = a^2 - b^2 = (a - b)(a + b)\)
Since 945 is positive, \(a - b\) and \(a + b\) must be either both positive or both negative.
Now factor 945:
\(945 = 3^3 * 5 * 7\)
So the number of positive divisors of 945 is:
\((3 + 1) * (1 + 1) * (1 + 1) = 16\)
For each positive divisor \(d\) of 945, we can set \(a - b = d\) and \(a + b = \frac{945}{d}\). Thus, there are 16 such ordered pairs with positive \(a - b\) and \(a + b\).
For example:
If \(a - b = 1\), then \(a + b = \frac{945}{1} = 945\).
Adding the two equations:
\(2a = 946\)
\(a = 473\)
Then \(b = 472\).
So this gives one ordered pair:
\((a, b) = (473, 472)\)
If \(a - b = 3\), then \(a + b = 315\).
Adding the two equations:
\(2a = 318\)
\(a = 159\)
Then \(b = 156\).
So this gives another ordered pair:
\((a, b) = (159, 156)\)
Also, reversing the factor values gives a different ordered pair.
For example:
If \(a - b = 945\), then \(a + b = 1\).
Adding the two equations:
\(2a = 946\)
\(a = 473\)
Then \(b = -472\).
So this gives:
\((a, b) = (473, -472)\)
Since 945 is odd, both \(d\) and \(\frac{945}{d}\) are odd, so the resulting values of \(a\) and \(b\) from each such pair will be integers. Thus, each positive divisor gives one ordered pair \((a, b)\). Since there are 16 positive divisors, we get 16 ordered pairs from positive factor pairs.
Similarly, the negatives \((-d, -\frac{945}{d})\) give another 16 ordered pairs of values for \(a - b\) and \(a + b\), with both values negative.
For example:
If \(a - b = -1\), then \(a + b = -945\).
Adding the two equations:
\(2a = -946\)
\(a = -473\)
Then \(b = -472\).
So the negative factor pairs give another 16 ordered pairs.
Therefore, the total number of ordered pairs \((a, b)\) is:
\(16 + 16 = 32\)
P.S. For reference, the 32 ordered pairs are:
\((473, 472), (159, 156), (97, 92), (71, 64), (57, 48), (39, 24), (33, 12), (31, 4)\)
\((31, -4), (33, -12), (39, -24), (57, -48), (71, -64), (97, -92), (159, -156), (473, -472)\)
\((-473, -472), (-159, -156), (-97, -92), (-71, -64), (-57, -48), (-39, -24), (-33, -12), (-31, -4)\)
\((-31, 4), (-33, 12), (-39, 24), (-57, 48), (-71, 64), (-97, 92), (-159, 156), (-473, 472)\)
Answer: D
