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Find the Pythagorean triple that has 97 as the length of the hypotenuse without using a computer or calculator. 

 Nov 4, 2023
 #1
avatar+1199 
0

Let (a,b,97) be the Pythagorean triple we are looking for. We know that a2+b2=972=9409. We also know that a and b must be relatively prime, since the greatest common divisor of the legs of a Pythagorean triple is always 1.

One way to find a primitive Pythagorean triple is to use the Pythagorean formula and experiment with different values for a and b. In this case, we can start by trying to find a value for a that is relatively close to the square root of 9409. We know that 8100​<9409​<10000​, so we can try a=90. Substituting into the Pythagorean formula, we get 902+b2=9409, so b2=9409−902=329. Since 329 is a prime number, we know that b=17. Therefore, the Pythagorean triple with 97 as the hypotenuse is (90,17,97).

 Nov 5, 2023
 #2
avatar+806 
+1

 

972 = 9409   (c)  

902 = 8100   (a or b)  

172 =   289   (b or a)  

 

a2 + b2 must = c2  but, as you can see from the above, it does not.  

 

The Pythagorean triple with 97 as the hypotenuse is not (90,17,97).  

 

I recommend always checking an answer, if possible, before posting.  

.

 Nov 5, 2023
 #3
avatar+397 
+1

Let, 

\(\displaystyle a = m^{2}-n^{2},\\\displaystyle b=2mn,\\ \displaystyle c=m^{2}+n^{2},\)

 

then a, b and c  form a Pythagorean triple.

 

\(\displaystyle a^{2}+b^{2}=(m^{2}-n^{2})^{2}+(2mn)^{2} \\ \displaystyle =m^{4}+n^{4}-2m^{2}n^{2}+4m^{2}n^{2} \\ \displaystyle =m^{4}+n^{4}+2m^{2}n^{2}=(m^{2}+n^{2})^{2} \\ \displaystyle = c^{2}.\)

 

What we are looking for then are values of m and n such that m^2 + n^2 = 97.

m^2  = 97 - n^2, so try

n =1 , m^2 = 97 - 1 = 96, m not an integer,

n = 2, m^2 = 97 - 4 = 93  mnai,

n = 3, m^2 = 97 - 9 = 88  mnai,

n = 4, m^2 = 97 - 16 = 81, m = 9, got it.

 

So, a = 9^2 - 4^2 = 81 - 16 = 65,

b = 2*9*4 = 72.

Check, (for silly mistakes), a^2 + b^2 = 65^2 + 72 ^2 = 4225 + 5184 = 9409 = 97^2. 

 Nov 6, 2023

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