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1,3,8 ...extend! (Posted on 2017-01-26) Difficulty: 3 of 5
The set (1,3,8) has this peculiar feature:

Multiply any two of these numbers together and add 1 and you'll always get a perfect square:

1+ 1 × 3 = 4
1 + 1 × 8 = 9
1 + 3 × 8 = 25

a. Add to the above set a 4th integer, thus enabling 3 additional equalities.
b. Can you show why no 5th member maintaining the above feature exists?

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution The number Comment 4 of 4 |
a. 120
1+(1*3)=4=2^2
1+(1*8)=9=3^2
1+(3*8)=25=5^2
1+(1*120)=121=11^2
1+(3*120)=361=19^2
1+(8*120)=961=31^2

b. ?


  Posted by Math Man on 2017-02-19 21:09:15
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