Question 1193263
<pre>
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 

{{{system(73a+70=x^2,73(a+1)+70=y^2)}}}

{{{system(73a=x^2-70,73a+73+70=y^2)}}}

{{{system(73a=x^2-70,73a+143=y^2)}}}

{{{system(73a=x^2-70,73a=y^2-143)}}}

{{{x^2-70=y^2-143}}}

{{{143-70=y^2-x^2}}}

{{{73=(y-x)(y+x)}}}

73 is a prime number and only has factors 1 and 73, so

{{{system(y-x=1,y+x=73)}}}

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

{{{system(73a=x^2-70,73a=y^2-143)}}}

{{{system(73a=36^2-70,73a=37^2-143)}}}

{{{system(73a=1226,73a=1226)}}}

{{{a=1226/73=16.79452055}}}

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</pre>