A number theory problem by Vilakshan Gupta

t suffices to prove $5\mid n,8\mid n$ .

Given that $2n+1=x^2, 3n+1=y^2$ , If we add the two equations we get: $5n+2=x^2+y^2\Rightarrow x^2+y^2\equiv 2\pmod 5$ , since the quadratic residues of mod $5$ are $0,1,4$ , we can check that only $x^2\equiv y^2\equiv 1$ works, therefore $2n\equiv 1-1\equiv 0\pmod 5\Rightarrow 5\mid n$

The same analysis can be done for $8$ , but we will multiply the second equation by $2$ first and then add:
$8n+3=x^2+2y^2\Rightarrow x^2+2y^2\equiv 3\pmod 8$ , since the quadratic residues of mod $8$ are $0,1,4$ , we can check that only $x^2\equiv y^2\equiv 1$ works again, which means $3n\equiv 1-1\equiv 0\pmod 8\Rightarrow 8\mod n$ and we are done.

Similar reasoning as @Md Zuhair , perhaps a little more step-by-step:

Considering that $2n+1=a^2$ while $3n+1=b^2$ for some integers $a$ and $b$ , we first make $n$ the subject of the equation by subtracting the former equation from the latter as follows:

$(3n+1)-(2n+1)=b^2-a^2$

$n=(b-a)(b+a)$ upon cancelling terms.

Two possibilities - either $a$ and $b$ have the same parity (i.e. both odd or both even), otherwise $a$ and $b$ have opposite parity (i.e. one is odd, the other is even).

Suppose we have the latter case where one is odd while the other is even. Then both $(a+b)$ and $(a-b)$ would be odd which makes $n$ odd. In other words, $n=2k+1$ for some non-negative integer $k$ . But we encounter a glaring problem as when we substitute $n=2k+1$ into $2n+1=a^2$ , we have $a^2=2(2k+1)+1=2\times2k+(2+1)=4k+3$ . Note that a perfect square only leaves a remainder of 0 or 1 (never 3) when divided by 4.

We thus reject the latter possibility and can safely assume that the former is true. In fact, $2n+1=a^2$ tells us $a$ must be odd so if $b$ has the same parity as $a$ , $b$ must be odd too. Letting $a=2h+1$ and $b=2k+1$ , $a+b=(2h+1)+(2k+1)=2h+2k+2=2(h+k+1)$ and $a-b=(2h+1)-(2k+1)=2h-2k=2(h-k)$ . Since $h$ and $k$ are both integers, it can be easily seen that both $(h+k)$ and $(h-k)$ must have the same parity (both even or both odd) which means $(h+k+1)$ and $(h-k)$ have opposite parity (one is odd, the other even). In other words, either of $(h+k+1)$ and $(h-k)$ is even or divisble by 2. So either of $a+b=2(h+k+1)$ or $a-b=2(h-k)$ is divisible by $2\times 2=4$ . The other is divisble by 2.

This tells us that $n=a^2-b^2$ must be divisible by $4\times 2=8$ .

Moreover, adding $2n+1=a^2$ to $3n+1=b^2$ gives us:

$(2n+1)+(3n+1)=a^2+b^2$

$(2+3)n+(1+1)=a^2+b^2$

$5n+2=a^2+b^2$

Note that a perfect square only leaves a remainder of 0, 1 or 4 when divided by 5. So for $a^2+b^2$ to leave a remainder of 2 when divided by 5, the only possibility is both $a^2$ and $b^2$ leave a remainder of 1 when divided by 5. Then $2n=a^2-1$ and $3n=b^2-1$ tell us that $2n$ and $3n$ are divisible by 5. Since 2 and 3 are coprime to 5, $n$ is divisible by 5 too.

Finally, the smallest $n=8\times5=\boxed{40}$

First, observe that the last digit of any perfect square is 1, 4, 5, 6, or 9.

Now consider n = 10k + m, m is 0-9. The last digit of 2n + 1 is 2m + 1 mod 10. So m is among 0, 2, 4, 5, 7, 9 if 2n + 1 is perfect square.

The last digit of 3n + 1 is 3m + 1 mod 10. This is when m is among 0, 1, 3, 5, 6, 8 if 3n + 1 is a perfect square.

So m is 0 or 5 if both cases are to be satisfied. A quick check among one or two digit numbers shows that 40 is the smallest, with 2n + 1 = 81 and 3n + 1 = 121.