55rd Ukrainian National Mathematical Olympiad - Fourth Round

For $a=1$ we have $p=1$ and $\frac{1}{2}(p^2+1)=1$ satisfies the problem.

For $a=2$ we have $p=3$ and $\frac{1}{2}(p^2+1)=5$ doesn't satisfy the problem.

Assume now $a\ge 3$. Let $\frac{1}{2}(p^2+1)=p_1^2$, then $p^2-2p_1^2=-1$. Hence: $2^{2a}-2^{a+1}+1-2p_1^2=-1$ or $2^{2a-1}-2^a=p_1^2-1$. So $2^a(2^{a-1}-1)=(p_1-1)(p_1+1)$. LHS is even, so as RHS. So $\gcd (p_1-1,p_1+1)=2$. So only the following cases are possible.

1) $p_1+1=2l$, and $kl=2^{a-1}-1$.

If $k\ge 2$ $p_1\ge 2^a+1$ and $l\ge 2^{a-1}+1$, a contradiction with $kl=2^{a-1}-1$.

If $k=1$ $p_1=2^{a-1}+1$ and $l=2^{a-2}+1$ and $kl=2^{a-2}+1=2^{a-1}-1$. Hence $a=3$.

2) $p_1-1=2k$, $p_1+1=2^{a-1}l$, and $kl=2^{a-1}-1$.

If $l\ge 2$ $p_1\ge 2^a-1$ and $k\ge 2^{a-1}-1$, so $2^{a-1}-1=kl\ge 2^a-2$, hence $a=1$.

If $l=1$ $p_1=2^{a-1}-1$ and $k=2^{a-2}-1$ and $kl=2^{a-2}-1=2^{a-1}-1$ - contradiction.

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Alternative solution.

Substitute $p=2^a-1$ into equality $\frac{1}{2}(p^2+1)=n^2$: $$(2^a-1{)}^2+1=2n^2.(\ast )$$ If $a=1$ we have $n=1$ satisfies the problem.

If $a>1$ we have that equality (*) becomes $2^{2a-1}-2^a+1=n^2$, so $n$ is odd. The last equality can be transformed to: $(2^{a-1}{)}^2+(2^{a-1}-1{)}^2=n^2$ or $$2^{2a-2}=n^2-(2^{a-1}-1{)}^2=(n-2^{a-1}+1)(n+2^{a-1}-1)\iff 2^{2a-2}=xy.$$ Also $x=n-2^{a-1}+1<y=n+2^{a-1}-1$. If $x,y$ are even, then they are powers of two, and also $y-x=2^a-2\equiv 2(\mathrm{mod}4)$. Then $x=2$, and $y=2^a=2^{2a-3}\Rightarrow$ the second solution is $a=3$.