Difference between revisions of "1993 AIME Problems/Problem 4"
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Square both sides of the first equation in order to get <math>bc</math> and <math>ad</math> terms, which we can plug <math>93</math> in for. | Square both sides of the first equation in order to get <math>bc</math> and <math>ad</math> terms, which we can plug <math>93</math> in for. | ||
− | <math>(a+d)^2 = (b+c)^2 \implies a^2 + 2ad + d^2 = b^2 + 2bc + c^2 \implies 2bc-2ad = a^2-b^2 + d^2-c^2 \implies 2(bc-ad) = (a-b)(a+b)+(d-c)(d+c)</math> | + | <math>(a+d)^2 = (b+c)^2 \implies a^2 + 2ad + d^2 = b^2 + 2bc + c^2 \implies 2bc-2ad = a^2-b^2 + d^2-c^2 \implies 2(bc-ad) = (a-b)(a+b) |
+ | +(d-c)(d+c)</math> | ||
We can plug <math>93</math> in for <math>bc - ad</math> to get <math>186</math> on the left side, and also observe that <math>a-b = c-d</math> after rearranging the first equation. Plug in <math>c-d</math> for <math>a-b</math>. | We can plug <math>93</math> in for <math>bc - ad</math> to get <math>186</math> on the left side, and also observe that <math>a-b = c-d</math> after rearranging the first equation. Plug in <math>c-d</math> for <math>a-b</math>. |
Revision as of 09:19, 18 June 2019
Problem
How many ordered four-tuples of integers with satisfy and ?
Contents
Solution
Solution 1
Let so . It follows that . Hence .
Solve them in tems of to get . The last two solutions don't follow , so we only need to consider the first two solutions.
The first solution gives us and , and the second one gives us .
So the total number of such four-tuples is .
Solution 2
Let and . From , .
Substituting , , and into , Hence, or .
For , we know that , so there are four-tuples. For , , and there are four-tuples. In total, we have four-tuples.
Solution 3
Square both sides of the first equation in order to get and terms, which we can plug in for.
We can plug in for to get on the left side, and also observe that after rearranging the first equation. Plug in for .
Now observe the possible factors of , which are . and must be factors of , and must be greater than .
work, and yields possible solutions. does not work, because if , then must differ by 2 as well, but an odd number can only result from two numbers of different parity. will be even, and will be even, so must be even. works, and yields possible solutions, while fails for the same reasoning above.
Thus, the answer is
Solution 4
Add the two conditions together to get . Rearranging and factorising with SFFT, . This implies that for every quadruple , we can replace , , etc. and this will still produce a valid quadruple. This means, that we can fix , and then just repeatedly add to get the other quadruples.
Now, our conditions are and . Replacing in the first equation, we get . Factorising again with SFFT gives . Since , we have two possible cases to consider.
Case 1: , . This produces the quadruple , which indeed works.
Case 2: , . This produces the quadruple , which indeed works.
Now, for case 1, we can add to each term exactly times (until we get the quadruple ), until we violate . This gives quadruples for case 1.
For case 2, we can add to each term exactly times (until we get the quadruple ). this gives quadruples for case 2.
In conclusion, having exhausted all cases, we can finish. There are hence possible quadruples.
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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