Proof using primes, divisibility, and sum of squares

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The discussion revolves around proving two mathematical statements involving prime numbers and sums of squares. For part a, the user attempts to show that if a prime p divides the sums of squares a² + b² and c² + d², then it also divides the difference a² - c². The user derives an equation involving b² - d² but is unsure how to proceed. The suggestion of looking for counterexamples is offered as a potential strategy to simplify the problem. The conversation emphasizes the challenge of proving these divisibility properties related to primes and squares.
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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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