- #36
roger
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matt grime said:Have you considered checking a couple of counter examples, with say, p=2 or p=3, or b=0?
Why would p=2 or 3 be counterexamples?
matt grime said:Have you considered checking a couple of counter examples, with say, p=2 or p=3, or b=0?
matt grime said:When p=3 it reduces to a+b. So you're honestly attempting to claim that a+b is always 0 mod 3? Nonsense. Just think about it for more than one second and stop wasting people's time.
roger said:cheers matt.
But why must a and b be <p for it to work?
roger said:a^3+b^5+c^7=d^11 iff a,b,c and d are natural numbers.
matt grime said:This makes no sense, again. You need to write things in sentences, and ones that make sense at that. We keep telling you this.
Are you asking to find the integer solutions to this?
roger said:a^3+b^5+c^7=d^11 iff a,b,c and d are natural numbers.
Is there a way to simplify this in order to find a,b,c and d?
and is there any significance in the powers being prime?
roger said:But the 'counter examples you gave were based on any value of a and b whereas afterwards I stated that they must be positive integers.
Seems I can't help myself. The point is that the counter examples all broke your initially unstated rules that 0<a<b<p. You asked why must a,b<p for this to work (which is a bad sentence by the way - use proper sentences). The answer is that they need not be, but that this condition, with the positivity condition will force it to work. Besides, you're just reducing everything mod p and so you only care about a,b mod p anyway.roger said:can someone else help me please?
The counter examples were all false since a and b must be positive integers.
roger said:The pattern is that a and p or b and p are coprime?
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