+142 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?
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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+129852 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
