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 | |
+---+---+---+

(In reply to

No Subject by Charlie)

..."Also it can be extended** two more rows**:...

Why **two**? It can be extended indefinitely in both directions, as long as the triplet of integers is ** t, 25-t, (12-t)^2-t.**

e.g. t=9: 9,16,-5; t=26: 26,-1,170; t=100: 100,-75; 7644.

Extra challenges(everybody, not just Ch):

1.Derive the the formula (**in bold **) for (a,b,c)- a+b=25

2,Derive a general formula for (a,b,c), any two add up to
a square, given a+b=(odd integer)^2.

BTW, Ch - 2nd row in the bottom table is erroneous (typo),

please edit for history sake,