Vtyjkyuk

Does there exist a perfect square of the form $5+40n$ for positive integers $n$? I know this is probably easy to show but any help is appreciated

What are the squares modulo $40$? Look in the OEIS to find http://oeis.org/A070462

$$0, 1, 4, 9, 16, 25, 36, 9, 24, 1, 20, 1, 24, 9, 36, 25, 16, 9, 4, 1, 0, 1, 4, 9, 16, ldots$$

This tells us that there are squares of the form $40n + 1$, $40n + 4$, $40n + 9$, etc. There's no $40n + 5$.

If there was, the square root would have to be an odd multiple of $5$. But we find that $5^2 equiv 25 pmod40$, $15^2 equiv 25 pmod40$ also, $25^2 equiv 25 pmod40$ and $35^2 equiv 25 pmod40$ also!

If such a perfect square existed that would imply that $a^2=5mod8$ has a solution which is not true (if you don't have much background just check it by hand).

If we write $40n+5 = a^2$ we get $a=5b$ so $8n+1 = 5b^2;;(*)$ and thus $5mid 8n+1$.

From here we get $$5mid 2(8n+1)-15n = n+2$$ so we can write $n+2=5k$. If we put $n=5k-2$ in to equation (*) we get: $$ 8k-3=b^2$$ Now $$b equiv_8 0,pm1,pm2pm3,4implies b^2 equiv_8 0,1,4notequiv -3$$

Let $40n+5=x^2$. Then for this to be true we need to find modular square roots of $5$ modulo $40$. That is
$$x^2equiv5pmod40tag1$$
Now if a congruence holds modulo $m$, it also holds modulo any divisor of $m$. Hence $(1)$ is equivalent to finding solutions to
$$x^2equiv5equiv0pmod5quadtextandquad x^2equiv5pmod8 tag2$$
We find $0$ to be a modular square root of $5$ modulo $5$, and to check for modular square roots of $5$ modulo $8$, we find the squares of the residues $pm1$, $pm2$, $pm3$, $pm4$ modulo $8$ give $0$, $1$, $4$, none of which is $equiv5pmod8$.

Hence $40n+5=x^2$ has no solutions in positive integers.