Mike D. answered • 08/25/23
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Let a, b be the integers, with a<2b and a^2 + b^2 = 925
You need to find two square numbers like 1^2, 2^2 .... that add up to 925.
The obvious solution is 25 and 900 giving a^2 = 25, so a=5 and b^2 = 900 so b = 30
So the integers are 5, and 30, and obviously 5 < 2.30
Mike D.
The . In the end 2.30 is multiplication
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08/26/23
Martin C.
tutor
There are two other solutions to this problem: 21 and 22, and 27 and 14. They can be found from Diophantus' theorem about the product of two numbers that are the sum of two squares itself being the sum of two squares. Here, 925 = 37 x 25 = (6^2 + 1^2) x (4^2 + 3^2). The theorem now says that the solutions are (6 x 4) + (3 x 1) and (6 x 3) + (4 x 1) for the two larger integers, and (6 x 3) - (4 x 1) and (6 x 4) - (3 x 1) for the smaller ones.
My guess is that Mike D's obvious solution of 5 and 30 was not one of the intended ones, as 30 is not less than twice 5, while the other two solutions do satisfy the extra condition.
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08/26/23
Arthur D.
tutor
I agree with Mike D. This is an algebra l course which would not contain the concept of Diophantus' Theorem. Since you have one equation with two variables it seems like a trial and error problem. Starting with 30^2 (31^2=961) and going down you come to 27^2 and then finally 22^2. Subtracting these numbers from 925 yields 25, 196 and 441-all perfect squares to give all three pairs of numbers satisfying the words of the problem.
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08/27/23
Brenda D.
tutor
Since there are several answers I too agree that Mike D’s result is one of them. I think it was made obvious to illustrate the point and serve the criteria, the student had to find the obvious and the not so obvious ones. One case of reasoning fits into the the Trial and Error; for positive integers, x is smaller than the next consecutive positive integer (x +1) so inherently x<2(x+1). Actually x^2 + (x+1)^2 = 925 will yield the positive 21. While SQRT(925 -x^2) allows the student to quickly test any number greater than 12. However, it is questionable whether the student’s Algebra class is into Quadratics, Pythagorean Triples just yet or Diophantus’ Theorem either so back to the Trial and Error. I wonder if the question was asked to see if the student would recognize the obvious Mike D’s approach and come up with others based on prior knowledge like time tables, how to square a number, and an understanding of square roots. With the exception 5^2 + 30^2 students should recognize the fact that the other eleven or ten perfect squares they should already know from multiplication tables (the ones from 0 to 12 or 0 to 10) do not sum to 925 or meet the requirements.
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09/02/23
James S.
tutor
The Pythagorean theorem and associated triples come up before Algebra I. Algebra I covers quadratic basics.
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09/04/23
Brenda D.
08/26/23