Question 1193263: prove that there do not exist three consecutive integer values of n for which 73n+70 is a perfect square. Found 2 solutions by Edwin McCravy, ikleyn:Answer by Edwin McCravy(20066) (Show Source):
We can do better than that. We can prove that there do not even exist
TWO consecutive integer values of n for which 73n+70 is a perfect square.
Assume for contradiction that there are two consecutive integers
'a', and 'a+1', and also integers 'x' and 'y' such that
73 is a prime number and only has factors 1 and 73, so
Adding those equations gives 2y = 74, or y = 37
Substituting 37-x=1, -x=-36, x=36
Now we substitute for x and y in
So 'a' cannot be an integer. And since we assumed that
'a' was an integer, we have reached a contradiction.
Since there are no TWO consecutive integer values for n
for which 73n+70 is a perfect square, there certainly are
no THREE such.
Edwin
You can put this solution on YOUR website! .
prove that there do not exist three consecutive integer values of n for which 73n+70 is a perfect square.
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To make a harmony from a disharmony, I would re-formulate the problem this way
Prove that there do not exist three consecutive integer values
of n such that the numbers 73n+70 all are perfect squares.
Let assume that there are (do exist) three consecutive integer values of n such that
three corresponding numbers 73n+70 are perfect squares.
Consider remainders of division these three integer numbers by 3.
Notice that 73n+70 gives the same remainder modulo 3 as the number (n+1).
So, the remainders of division the numbers 73n+70 by 3 are three numbers 0, 1 and 2 in some order.
In this chain of arguments, the fact which is important for us is that the remainder 2 with necessity
is one of the three remainders of (73n+70) modulo 3.
But no one perfect square may have remainder 2 when divided by 3: the possible remainders are only numbers 0 or 1.
