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In certain right triangles, $\sqrt{h^2-a^2} = 24$, where $h$ represents the hypotenuse and $a$ is the length of one of the legs. Find all possible ordered pairs $(h,a)$ where $h,a \in \mathbb{N}$.

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    Did you find any at all? – 2017-01-26
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    hint: factorise the left-hand side and write down 576 as product of two integer factors in every possible way. – 2017-01-26

2 Answers 2

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$$24^2=h^2-a^2=(h-a)(h+a).$$

Check all ways to factor $24^2$, and solve for $h$ and $a$. For example, $24=2 \cdot 288$ so solve $h-a=2$ and $h+a=288$ to get one solution.

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    Don't you mean $24^2 = 576 = (h-a)(h+a)$? – 2017-01-26
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    @fleablood sorry, thanks – 2017-01-26
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$\sqrt{h^2 - a^2} = 24$

$h^2 - a^2 = 24^2$

$(h-a)(h+a) = 2^6*3^2$

So $h-a = 2^j*3^k$ and $h+a = 2^{6-j}*3{2-k}$ and $2h = 2^j*3^k+ 2^{6-j}*3{2-k}$ and $2a = 2^{6-j}*3{2-k}-2^j*3^k$

So $h = 2^{j-1}3^k+2^{6-j-1}*3^{2-k}$ and $a = 2^{6-j -1}3^{2-k}- 2^{j-1}3^k$.

So $1 \le j \le 5$

Case 1: $j = 1$ then

$h = 3^k + 16*3^{2-k}$ and $a = 16*3^{2-k} - 3^k$

so for $k = 0, 1,2$

$h = 1 + 16*9;a = 16*9 -1$ or $h = 3 + 16*3; a = 16*3 - 3$ or $h= 9 +16=25; a = 16-9 = 7$

Case 2: $j =2$ then

$h = 2*3^k + 8*3^{2-k}$ and $a = 8*3^{2-k} - 2*3^k$

so for $k = 0, 1,2$

$h = 2 + 8*9;a = 8*9 -2$ or $h = 6 + 8*3; a = 8*3 - 6$ or $h= 18 +8=26; a = 8-18 = -10XXX$

Case 3: $j =3$ then

$h = 3^k*4 + 4*3^{2-k}$ and $a = 4*3^{2-k} - 4*3^k$

so for $k = 0, 1,2$

$h = 4 + 4*9=40;a = 4*9 -4=32$ or $h = 12 + 12=24; a = 12 - 12=0XXX$ or $h= 36 +4=26; a = 4-36 = -32XXX$

Case 4: $j =4$ then

$h = 3^k*8 + 2*3^{2-k}$ and $a = 2*3^{2-k} - 8*3^k$

so for $k = 0, 1,2$

$h = 8 + 2*9=26;a = 2*9 -8=11$ or $h = 24 + 6=30; a = 6-24=-18XXX$ or $h= 72 +2=74; a = 2-74 = -72XXX$

Case 5: $j= 5$ then

$h = 3^k*16 + 3^{2-k}$ and $a =3^{2-k} - 3^k*16$

So $h=16 + 9;a = 9 -16 XXX; etc.$.

So solutions are:

$h =145;a = 144$ or $h = 51; a = 45$ or $h= 25; a = 7$

$h = 74;a = 70$ or $h = 30; a = 18$

$h =40;a =32$

$h =26;a =11$