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If 3n + 1 is a perfect square, show that n + 1 is the sum of three perfect squares.

 Nov 27, 2015

Best Answer 

 #1
avatar+26396 
+35

If 3n + 1 is a perfect square, show that n + 1 is the sum of three perfect squares.

 

3n + 1 is a perfect square:

3n+1=a23n=a21n=a213n+1=a213+1n+1=a21+33n+1=a2+23

 

Because a2+23 is a integer then a2+2 is divisible by 3, then a2 is not divisible by 3  and also a is not divisible by 3,

because if 3 is not a prime factor in a2 a perfect spuare, then 3 is not a prime factor in  a

 

Two numbers are not divisible by 3. It is  3b+1 and 3b+2

 

1. We substitute a=3b+1

n+1=a2+23substitute  a=3b+1n+1=(3b+1)2+23n+1=9b2+6b+1+23n+1=9b2+6b+33n+1=3b2+2b+1n+1=b2+b2+b2+2b+1n+1=b2+b2+(b+1)2

 

So n + 1 is the sum of three perfect squares and b=a13.

 

2. We substitute a=3b+2

n+1=a2+23substitute  a=3b+2n+1=(3b+2)2+23n+1=9b2+12b+4+23n+1=9b2+12b+63n+1=3b2+4b+2n+1=b2+b2+b2+2b+2b+1+1n+1=b2+b2+2b+1+b2+2b+1n+1=b2+(b+1)2+(b+1)2

 

So n + 1 is the sum of three perfect squares and b=a23.

 

Example 1:

a=73n+1=a2=72n=a213=4913=16n+1=16+1=1717=b2+b2+(b+1)2a=3b+1b=a13=713=217=22+22+32=4+4+9

 

Example 2:

a=83n+1=a2=82n=a213=6413=21n+1=21+1=2222=b2+(b+1)2+(b+1)2a=3b+2b=a23=823=222=22+32+32=4+9+9

 

 

laugh

 Nov 27, 2015
edited by heureka  Nov 27, 2015
edited by heureka  Nov 27, 2015
 #1
avatar+26396 
+35
Best Answer

If 3n + 1 is a perfect square, show that n + 1 is the sum of three perfect squares.

 

3n + 1 is a perfect square:

3n+1=a23n=a21n=a213n+1=a213+1n+1=a21+33n+1=a2+23

 

Because a2+23 is a integer then a2+2 is divisible by 3, then a2 is not divisible by 3  and also a is not divisible by 3,

because if 3 is not a prime factor in a2 a perfect spuare, then 3 is not a prime factor in  a

 

Two numbers are not divisible by 3. It is  3b+1 and 3b+2

 

1. We substitute a=3b+1

n+1=a2+23substitute  a=3b+1n+1=(3b+1)2+23n+1=9b2+6b+1+23n+1=9b2+6b+33n+1=3b2+2b+1n+1=b2+b2+b2+2b+1n+1=b2+b2+(b+1)2

 

So n + 1 is the sum of three perfect squares and b=a13.

 

2. We substitute a=3b+2

n+1=a2+23substitute  a=3b+2n+1=(3b+2)2+23n+1=9b2+12b+4+23n+1=9b2+12b+63n+1=3b2+4b+2n+1=b2+b2+b2+2b+2b+1+1n+1=b2+b2+2b+1+b2+2b+1n+1=b2+(b+1)2+(b+1)2

 

So n + 1 is the sum of three perfect squares and b=a23.

 

Example 1:

a=73n+1=a2=72n=a213=4913=16n+1=16+1=1717=b2+b2+(b+1)2a=3b+1b=a13=713=217=22+22+32=4+4+9

 

Example 2:

a=83n+1=a2=82n=a213=6413=21n+1=21+1=2222=b2+(b+1)2+(b+1)2a=3b+2b=a23=823=222=22+32+32=4+9+9

 

 

laugh

heureka Nov 27, 2015
edited by heureka  Nov 27, 2015
edited by heureka  Nov 27, 2015
 #2
avatar+130466 
+5

Very nice, heureka.......!!!!

 

 

cool cool cool

 Nov 27, 2015

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