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 Adjacent Square Sum II (Posted on 2013-06-10)
In the six rows of numbers below, each of the pairs adds up to 25. Now 25 happens to be a perfect square.

Fill in the blanks with a third number (a different number in each row) so that the sums of any two numbers on any row is a perfect square.
```+---+---+---+
| 1 |24 |   |
+---+---+---+
| 2 |23 |   |
+---+---+---+
| 3 |22 |   |
+---+---+---+
| 4 |21 |   |
+---+---+---+
| 5 |20 |   |
+---+---+---+
| 6 |19 |   |
+---+---+---+```

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 More generalizations of the algebra | Comment 4 of 5 |
(In reply to The algebra by Jer)

It's not all Pythagorean triple, just the ones like (5,12,13) and (7,40,41)...(2a+1, 2a²+2a, 2a²+2a+1)

For pairs that add to (2a+1)², call them n and (2a+1)²-n
you can add the quantity n²+n(-4a²-4a-1)+(4a^4+8a³+4a²)
which yields the squares
n²+n(-4a²-4a)+(4a^4+8a³+4a²) = (n-(2a²+2a))²
and
n²+n(-4a²-4a-2)+(4a^4+8a³+8a²+4a+1) = (n-(2a²+2a+1))²
 Posted by Jer on 2013-06-10 15:28:26

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