Proof using primes, divisibility, and sum of squares

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SUMMARY

The discussion revolves around proving two mathematical statements involving prime numbers and sums of squares. In Part a, the claim is that if a prime number p divides both (a² + b²) and (c² + d²), then it also divides (a² - c²). The user demonstrates an initial approach using integer substitutions and manipulations, ultimately concluding that p must divide (b² - d²). In Part b, the assertion is that p also divides (a² + c²) under the same conditions, which the user anticipates will follow a similar logic.

PREREQUISITES
  • Understanding of prime numbers and their properties
  • Familiarity with modular arithmetic
  • Knowledge of algebraic manipulation and substitutions
  • Basic concepts of divisibility and integer equations
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  • Research the properties of prime numbers in modular arithmetic
  • Study the implications of sums of squares in number theory
  • Explore counterexamples in divisibility proofs
  • Learn about algebraic identities related to squares and their differences
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Mathematics students, educators, and anyone interested in number theory, particularly those working on proofs involving primes and divisibility.

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Homework Statement


I have to prove or disprove the following:

Part a) If p is prime and p | (a2 + b2) and p | (c2 + d2), then p | (a2 - c2)

Part b) f p is prime and p | (a2 + b2) and p | (c2 + d2), then p | (a2 + c2)

Homework Equations





The Attempt at a Solution



Part a)

Since p | (a2 + b2), we have that a2 + b2 = pk, for some integer k.
Since p | (c2 + d2), we have that c2 + d2 = pt, for some integer t.

Suppose p | (a2 - c2), then we have that a2 - c2 = pr, for some integer r.

By solving for a2 and c2 in the above equations, and substitution we have that
pk - b2 - (pt - d2) = pr
pk - pt - b2 + d2 = pr
pk - pt - pr = b2 - d2
p (k - t - r) = b2 - d2
So p | (b2 - d2)

I don't know where to go from here.

I figure that once I figure out how to do this part, the second part should be very similar.
Any help would be greatly appreciated. I'm going CRAZY trying to figure this out...
 
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Did you try looking for counterexamples before going CRAZY? That's always a good idea.
 

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