12 multiple-choice questions, progressively harder.
Which ordered triple does the parameter pair (12,5)(12,5)(12,5) give when the square difference is written first?
Solution
Correct answer: B
Square both parameters before adding or subtracting. Use the doubled product for the other leg.
The pair (16,10)(16,10)(16,10) generates a scaled triple. Which triple remains after division by its GCF? Use square-difference leg first.
Correct answer: C
The parameters share 222, and dividing both by 222 gives the coprime pair (8,5)(8,5)(8,5) of opposite parity. Each original output is 444 times its new partner.
For the generated triple from parameter pair (17,9)(17,9)(17,9), what is the GCF of the three entries?
Correct answer: A
The parameter squares are odd, so all three outputs are even. The sum of two odd squares leaves remainder 222 when divided by 444.
No odd prime divides both the square sum and difference because it would divide both 171717 and 999. Therefore 222 is the full GCF.
The diagram shows a primitive triple. Which pair (m,n)(m,n)(m,n) generated its sides?
Correct answer: D
Use the odd leg 161161161 with the hypotenuse 289289289, even though the horizontal leg is longer.
The positive roots are 151515 and 888.
The lengths 696969, 792792792, and 795795795 form a right triangle. What are the two parameter squares of its primitive triple, larger first?
The side lengths have GCF 333. Dividing gives (23,264,265)(23,264,265)(23,264,265).
Which pair lies in the domain m>n>0m>n>0m>n>0 and produces a scaled triple?
Only (6,2)(6,2)(6,2) meets both strict inequalities. Its shared parameter factor 222 contributes a square factor to all the outputs.
The product of 121212 and 272727 is a perfect square. Why does this example fall outside the claim that two coprime factors of a square must be squares?
The two factors share 333. The exponent of 333 in the product is even, although the exponent in each factor is odd.
An odd prime divides both the square-sum output and the square-difference output. What must it also divide?
The prime divides the sum and difference of those outputs.
As it is odd, it must divide both parameter squares and therefore both parameters.
A generated triple has c−a=200c-a=200c−a=200, where aaa is the square-difference leg. The parameters differ by 111. What is its even leg?
Subtract the square-difference formula from the square-sum formula to isolate the smaller parameter square.
Then use the doubled product.
The coprime pairs (18,1)(18,1)(18,1) and (17,6)(17,6)(17,6) have opposite parity. They both give hypotenuse 325325325. What follows?
Both outputs are primitive. Their odd legs differ, so the two triangles have different side lengths.
In the reverse proof, suppose a prime divides both the half-sum and the half-difference of the hypotenuse and odd leg. Which consequence rules this out for a primitive triple?
The prime would divide the sum ccc and the difference aaa. It would then divide the even leg square.
A prime factor of b2b^2b2 also divides bbb. A shared prime among all three sides contradicts GCF 111.
A valid pair has m>n>0m>n>0m>n>0. What happens if their positions are interchanged in all three formulas?
The product and the sum are unchanged by switching order. The difference reverses.
It becomes negative, so the interchanged inputs do not give three positive lengths.
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