Question 1193263
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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


<pre>
            Prove that there do not exist three consecutive integer values 
            of n such that the numbers  73n+70  all are perfect squares.
</pre>


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.


This contradiction PROVES the statement.


Solved.