Prove that there are no nonzero integers [tex]$a, b, c$[/tex] such that [tex]$a^2+b^2=3 c^2$[/tex]. (Hint: By studying the equation [tex]$[a]_4^2+[b]_4^2=3[c]_4^2$[/tex] in [tex]$Z / 4 Z$[/tex], show that [tex]$a, b, c$[/tex] would all have to be even. Letting [tex]$a=2 k, b=2 \ell, c=2 m$[/tex], you would have [tex]$k^2+\ell^2=3 m^2$[/tex]. What's wrong with that?)
Answer (2)
Analyze the equation a 2 + b 2 = 3 c 2 modulo 4.
Determine that a , b , c must all be even.
Substitute a = 2 k , b = 2 l , c = 2 m into the equation, obtaining k 2 + l 2 = 3 m 2 .
Apply the method of infinite descent to conclude that the only solution is a = b = c = 0 , thus there are no nonzero integer solutions: no nonzero integers a , b , c .
Explanation
Problem Analysis We are asked to prove that there are no nonzero integers a , b , c such that a 2 + b 2 = 3 c 2 . We will analyze this equation modulo 4, as suggested by the hint.
Possible values of squares modulo 4 First, let's consider the possible values of x 2 ( mod 4 ) for any integer x . From the calculation tool, we know that:
0 2 ≡ 0 ( mod 4 ) 1 2 ≡ 1 ( mod 4 ) 2 2 ≡ 0 ( mod 4 ) 3 2 ≡ 1 ( mod 4 )
So, the only possible values for x 2 ( mod 4 ) are 0 and 1.
Possible values of left side modulo 4 Now, let's analyze the equation a 2 + b 2 = 3 c 2 ( mod 4 ) . We know that a 2 ( mod 4 ) and b 2 ( mod 4 ) can only be 0 or 1. Thus, a 2 + b 2 ( mod 4 ) can be 0 + 0 = 0 , 0 + 1 = 1 , or 1 + 1 = 2 . So, a 2 + b 2 ( mod 4 ) can be 0, 1, or 2.
Possible values of right side modulo 4 Next, let's consider 3 c 2 ( mod 4 ) . Since c 2 ( mod 4 ) can only be 0 or 1, 3 c 2 ( mod 4 ) can be 3 ⋅ 0 = 0 or 3 ⋅ 1 = 3 . So, 3 c 2 ( mod 4 ) can be 0 or 3.
Deduction that a, b, c are even Comparing the possible values of a 2 + b 2 ( mod 4 ) and 3 c 2 ( mod 4 ) , we see that the only common value is 0. This means that a 2 + b 2 ≡ 0 ( mod 4 ) and 3 c 2 ≡ 0 ( mod 4 ) . For a 2 + b 2 ≡ 0 ( mod 4 ) , we must have a 2 ≡ 0 ( mod 4 ) and b 2 ≡ 0 ( mod 4 ) , which implies that a ≡ 0 ( mod 2 ) and b ≡ 0 ( mod 2 ) . For 3 c 2 ≡ 0 ( mod 4 ) , we must have c 2 ≡ 0 ( mod 4 ) , which implies that c ≡ 0 ( mod 2 ) . Therefore, a , b , c must all be even.
Substituting a=2k, b=2l, c=2m Since a , b , c are all even, we can write a = 2 k , b = 2 l , c = 2 m for some integers k , l , m . Substituting these into the original equation, we get ( 2 k ) 2 + ( 2 l ) 2 = 3 ( 2 m ) 2 , which simplifies to 4 k 2 + 4 l 2 = 12 m 2 . Dividing by 4, we get k 2 + l 2 = 3 m 2 .
Infinite Descent Argument Notice that the new equation k 2 + l 2 = 3 m 2 has the same form as the original equation a 2 + b 2 = 3 c 2 . This means that k , l , m must also be even. We can repeat this process indefinitely, which implies that a , b , c are divisible by arbitrarily large powers of 2. The only integers that satisfy this condition are a = b = c = 0 .
Final Answer Since we are given that a , b , c are nonzero integers, we have reached a contradiction. Therefore, there are no nonzero integers a , b , c such that a 2 + b 2 = 3 c 2 .
Examples
Consider the equation a 2 + b 2 = 3 c 2 . This equation has no solutions in nonzero integers. This concept is related to quadratic forms and Diophantine equations, which have applications in cryptography and coding theory. For example, the impossibility of certain equations can be used to design secure encryption algorithms or to prove the security of existing ones. Understanding the properties of such equations helps in creating robust systems that are resistant to attacks.
By analyzing the equation a 2 + b 2 = 3 c 2 modulo 4, we find that a , b , and c must all be even, leading to a recursive set of equations which must ultimately yield zero. Hence, there are no nonzero integer solutions for the equation. The final conclusion is that no nonzero integers a , b , c satisfy the equation.
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