Answer:
(8, 15, 17); (9, 12, 15); (15; 20; 25)
Step-by-step explanation:
If
[tex]a=m^2-n^2;\ b=2mn;\ c=m^2+n^2[/tex]
for m > n, then a, b, c make a Pythagorean triplet.
[tex]m^2-n^2=15\to(m-n)(m+n)=15\\\\(m-n)(m+n)=(3)(5)\to m-n=3\ \wedge\ m+n=5[/tex]
We have the system of equations:
[tex]\underline{+\left\{\begin{array}{ccc}m-n=3\\m+n=5\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad2m=8\qquad\text{divide both sides by 2}\\.\qquad\boxed{m=4}[/tex]
Substitute to the second equation:
[tex]4+n=5\qquad\text{subtract 4 from both sides}\\\boxed{n=1[/tex]
Therefore we have:
[tex]a=4^2-1^2=16-1=15\\b=2(4)(1)=8\\c=4^2+1^2=16+1=17[/tex]
[tex]2mn=15[/tex] it's impossible, because 15 is not an even number.
[tex]m^2+n^2=15[/tex]
Let's consider all possible sums of two numbers resulting in 15.
We will check which of the numbers are perfect squares.
1 + 14
2 + 13
3 + 12
4 + 11
5 + 10
6 + 9
7 + 8
(Bold not perfect squares)
There are no two perfect squares among the listed pairs of numbers.
Other:
15, 112, 113
We know the Egyptian triangle with sides of length 3, 4, 5.
By modifying this Pythagorean triplet by multiplying by 3 we get:
(3)(3) = 9; (3)(4) = 12; (3)(5) = 15
By modifying this Pythagorean triplet by multiplying by 5 we get:
(5)(3) = 15; (5)(4) = 20; (5)(5) = 25
