Perfect Square Number, prime I want to show that $p^{p+1}+(p+1)^p$ is not perfect square number. ($p$ is prime number)
This is what I do
Assume $p^{p+1}+(p+1)^p=k^2$.
$(p+1)^p=(k+p^{\frac {p+1} 2})(k-p^{\frac {p+1} 2})$.
Let $2ab=p+1, gcd(a,b)=1$.
$k+p^{\frac {p+1} 2}=2^{p-1}a^{p}, k-p^{\frac {p+1} 2}=2b^p$.
Therefore $p^{\frac {p+1} 2}=|2^{p-2}a^p-b^p|$.
By Fermat's little theorem, $a\equiv 2b \pmod{p}$, and $2ab=p+1$, so $a=2b$. In this case I can show contradiction.
but, if $a=1$ I don't know how show contradiction.
 A: Your proof is quite faulty (mathematically speaking, you omitted 2-3 cases i think), so I will reweite it entirely.
$$p^{p+1}+(p+1)^p=k^2\Leftrightarrow (p+1)^p=\big(k-p^{\frac{p+1}{2}}\big)\big(k+p^{\frac{p+1}{2}}\big)$$
Notice that $\text{gcd}(p+1;p)=1$ and $\text{gcd}\big(k-p^{\frac{p+1}{2}};k+p^{\frac{p+1}{2}}\big)|2\cdotp^{\frac{p+1}{2}}$, thus leading to $$\text{gcd}\big(k-p^{\frac{p+1}{2}};k+p^{\frac{p+1}{2}}\big)|2$$
$(p+1)^p$ is even so at least one of $k-p^{\frac{p+1}{2}}$ and $k+p^{\frac{p+1}{2}}$ is even, thus $$\text{gcd}\big(k-p^{\frac{p+1}{2}};k+p^{\frac{p+1}{2}}\big)=2$$
Note that case $p=2$ does not work, so from now on we will only work with $p\geq 3$
We can now safely assume that we have $2$ cases:
$1)$ $k-p^{\frac{p+1}{2}}=2\cdot a^p$ and $k+p^{\frac{p+1}{2}}=2^{p-1}\cdot b^p$ with $2ab=p+1$
$2)$ $k-p^{\frac{p+1}{2}}=2^{p-1}\cdot a^p$ and $k+p^{\frac{p+1}{2}}=2\cdot b^p$ with $2ab=p+1$
(notice that $a$ abd $b$ are not divisible by $p$)

Case $1)$: Using Fermat's little theorem, we get $k\equiv2a\pmod{p}$ and $k\equiv b\pmod{p}$ so $2a\equiv b\pmod{p}$ but $2ab\equiv 1\pmod{p}$ so either $b=1$ and $a=\frac{p+1}{2}$ or $b=p-1$ and $a=\frac{p+1}{2(p-1)}$.
Clearly when $a=\frac{p+1}{2(p-1)}$ we get a contradiction (unless $p=3$, which is a special case, we can easily see that $4^3+3^4$ is not a perfect square)
So $b=1$ and $a=\frac{p+1}{2}$ But $k=2^{p-1}b^p-p^{\frac{p+1}{2}}$ so if $b=1$ then $2^{p-1}> p^{\frac{p+1}{2}}$ which is clearly false for all primes $p$.
So we only get contradictions in this case.

Case $2):$ Again, using the same method we get $2b\equiv a\pmod{p}$ so with the same reasoning $a=1$ and $b=\frac{p+1}{2}$
We know $k=2\cdot a^p+p^{\frac{p+1}{2}}=2^{p-1}\cdot b^p-p^{\frac{p+1}{2}}$ so we get $$2+p^{\frac{p+1}{2}}=2^{p-1}\cdot \big(\frac{p+1}{2}\big)^p-p^{\frac{p+1}{2}}$$ which can we rewritten as $$2(1+p^{\frac{p+1}{2}})=2^{p-1}\cdot \big(\frac{p+1}{2}\big)^p$$
We assumed $p\geq 3$ so $p\equiv 1$ or $3\pmod{4}$
If $p\equiv 1\pmod{4}$, $1+p^{\frac{p+1}{2}}\equiv 2\pmod{4}$ so $$v_2(2(1+p^{\frac{p+1}{2}})=2$$ but $$v_2(2(1+p^{\frac{p+1}{2}})=v_2(2^{p-1}\cdot \big(\frac{p+1}{2}\big)^p)\geq p-1$$ giving us $$3\geq p$$ so contradiction; we assumed $p\equiv 1\pmod{4}$
If $p\equiv 3\pmod{4}$ then $\frac{p+1}{2}$ is even, so $p^\frac{p+1}{2}\equiv 1\pmod{4}$ (any odd perfect square is $1\pmod{4}$) so we have again that $1+p^{\frac{p+1}{2}}\equiv 2\pmod{4}$ so $$v_2(2(1+p^{\frac{p+1}{2}})=2$$ giving us the same result, $3\geq p$, which would lead to $p=3$, which we established is a contradiction.

Thus, there is no prime $p$ for which $p^{p+1}+(p+1)^p$ is a perfect square.
