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Weird question help

So my friend gave me this problem:
2^a + 3^b = c^2 where a, b and c are all integers. I have got one solution where a=4, b= 2 and c=5 but I cant find any others. If this is the only solution I also have to prove why but have no idea how to do this.
Any help would be great
@TeeEm ("Miss" :P )
Reply 2
Avatar for Zacken
Zacken
22
Original post by Quantum42
So my friend gave me this problem:
2^a + 3^b = c^2 where a, b and c are all integers. I have got one solution where a=4, b= 2 and c=5 but I cant find any others. If this is the only solution I also have to prove why but have no idea how to do this.
Any help would be great


Remember that cc is squared, so you really mean c=±5c = \pm 5.

Also, what about a=0,b=1,c=±2a=0, b=1, c = \pm 2? Or a=3,b=0,c=±3a=3, b=0, c= \pm 3?
Reply 3
Avatar for TeeEm
TeeEm
19
Original post by FrenchUnicorn
@TeeEm ("Miss" :P )


sorry but I am teaching
Original post by TeeEm
sorry but I am teaching


PRSOM *so frustrating*
Reply 5
Avatar for Zacken
Zacken
22
Original post by Quantum42
So my friend gave me this problem:
2^a + 3^b = c^2 where a, b and c are all integers. I have got one solution where a=4, b= 2 and c=5 but I cant find any others. If this is the only solution I also have to prove why but have no idea how to do this.
Any help would be great


Some initial considerations, you want 2a+3b2^a + 3^b to be a square number. You also know that 22 and 33 are primes.
Reply 6
Avatar for Quantum42
Quantum42
OP
14
Original post by Zacken
Some initial considerations, you want 2a+3b2^a + 3^b to be a square number. You also know that 22 and 33 are primes.


Sorry still really confused as to how that helps

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