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Right \$\triangle ABC\$ has hypotenuse \$AB\$. Square \$BCDE\$ has \$BC\$ as one of its sides. Suppose that the area of \$BCDE\$ is a prime number. If \$AB\$ and \$AC\$ are each integers less than \$20\$, how many possibilities are there for their lengths?

Apr 17, 2019

#1
+1

I think there's only one possibility.  If the area of that square is a prime number, then either its width or its height has to be 1.  Let's say its width.  Any value for the width other than 1, when multiplied by the height will give a product that isn't prime.

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Apr 17, 2019
edited by Guest  Apr 17, 2019
edited by Guest  Apr 17, 2019
#2
0

Not sure if you know this, but the side lengths of a square are the same. So if the width of the square is 1, the area is 1 which is not prime. I'm pretty sure that this means there is no possibilities.

Guest Apr 17, 2019
#3
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Wait, never mind. There are possibilities. As long as AB-AC is prime, then the area of BCDE is prime.

Guest Apr 17, 2019
#4
+3

I believe this is correct

AB is the hypotenuse, so  AC  and AB are legs

So

BC^2  =  AB^2  - AC^2  =   area of  BCDE

BC^2  =  (AB + AC) (AB - AC)

If we let (AB - AC)  =  1......then  BC^2  will have only one prime factor, (AB + AC)

So  we have the following possibilites

AB  AC     BC   BC^2

19  18     √37     37

16  15     √31     31

15  14     √29     29

12  11     √23     23

10   9      √19    19

9     8       √17   17

7    6        √13   13

6    5        √11    11

4   3         √7      7

3   2         √5      5

2   1         √3      3   Apr 17, 2019