Factoring the left hand side, we know $(a-1)a(a+1) = 2b^2.$ We want, in other words, three consecutive positive integers that multiply to be twice a square number.

Suppose $a$ is even. Then $a-1,$ $a,$ and $a+1$ do not share any common factors. Hence, $a-1$ and $a+1$ must both be squares. Then $a-1$ and $a+1$ are odd squares that are 2 apart, but the smallest difference between positive squares is $2^2 - 1^2 = 3.$

Suppose $a$ is odd. Then $a-1$ and $a+1$ are both even. Apart from the prime factor of 2, these 3 numbers do not share any common factors. Then $a-1$ and $a+1$ can be written for some positive integers $x$ and $z$ as $2x^2$ and $2z^2$ which are 2 apart. However, the smallest difference between the double of positive squares is $2(2^2) - 2(1^2) = 6 .$

Since $a$ can be neither even nor odd, there is no solution for which $a$ and $b$ are both positive integers.

$\begin{aligned} 2b^2 &= a\left(a^2-1\right)\\ \Rightarrow \sqrt{2}b &= \sqrt{a\left(a^2-1\right)}\\ &= \sqrt{a-1}\sqrt{a}\sqrt{a+1} \end{aligned}$

This can only hold if one of the 3 consecutive integers, $a-1,a, a+1$ , is equal to $2A^2, A\in Z^+$ , and the remaining two are square numbers. But there is no gap between two consecutive square numbers smaller than $2^2-1^2=3$ , so this is not possible.

Observe that $a(a^2-1)=(a-1)a(a+1)$ . Hence $b^2 = \frac{1}{2}(a-1)a(a+1)$ . But the product of three consecutive numbers can never be a perfect square, hence $b^2 = \frac{1}{2}(a-1)a(a+1)$ won't be as well. (A generalization of this can be found in the paper "The product of consecutive integers is never a power" by Erdős and Selfridge)

I substituted k + 1 = a. Now the problem can be stated k(k+1)(k+2) = 2b^2. Dividing by 2 creates k(k+1)(k/2 + 1) = b^2. Since k and k+1 are consecutive integers, they would be relatively prime. This means that k/2 + 1 = k(k +1). This equation can be written as k + 2 = 2k^2 + 2k. In standard form set equal to zero, you get 2k^2 + k - 2 = 0. The discriminant of the equation is 5, which isn't a perfect square. This means k is not an integer, which contradicts a given condition for the problem.

In the range from the first positive integer 1 to +Infinity both functions are monotonically increasing.

f(a)= $a(a^{2}-1)$ starts at 0 and increases faster being a 3rd degree polynomial compared to 2nd degree polynomial

f(b)= $2b^{2}$ which starts at 2 and increases slower.

Therefore, for the proof, it suffices to show that the single crossover point is not an integer.

Simple check for function output for a=1,2,3 and b=1,2,3 shows that the crossover happens at a non-integer point between 2 and 3.

f(a) = 0, 6, 24, ...

f(b) = 2, 8, 18, ...

Practice my number theory solving skill:

I have modified my solution as below:

When $a\equiv 0\pmod2$ , let $a=2k$ , so $k(2k-1)(2k+1)=b^2$ , because $\gcd(k,2k-1)=\gcd(k,-1)=1\Bigg| \gcd(k,2k+1)=\gcd(k,1)=1\Bigg| \gcd(2k-1,2k+1)=\gcd(2k-1,2)=\gcd(-1,2)=1$ , so integers in the set $\{k,2k-1,2k+1\}$ are pairwise coprime, thus we can conclude that each can be expressed as square, but this leads to $\begin{cases}p^2=2k-1\\q^2=2k+1\end{cases}$ for some positive integers $p,q$ , hence $(q+p)(q-p)=2$ , but because $q+p\equiv q-p\pmod2$ so the product is either odd or multiple of $4$ , contradiction.

Now consider $a\equiv1\pmod4$ , let $a=4k+1$ , we can get $k(4k+1)(2k+1)=\bigg(\frac{b}{2}\bigg)^2$ , now again consider are them coprime, and the fact is indeed true: $\gcd(k,4k+1)=\gcd(k,1)=1\Bigg|\gcd(k,2k+1)=\gcd(k,1)=1\Bigg|\gcd(4k+1,2k+1)=\gcd(-1,2k+1)=1$ , so the integer in the set $\{k,4k+1,2k+1\}$ are pairwise coprime, therefore each of them can again be square, but this leads to $\begin{cases}p^2=4k+1\\q^2=k\end{cases}$ for some positive integers $p,q$ . Thus $4q^2+1=p^2$ , $(p+2q)(p-2q)=1$ , own a solution $p=\pm1, q=0$ . All of these making $a=1,b=0$ , which isn't a suitable answer due to question need, contradiction.

Finally, consider $a\equiv3\pmod4$ , let $a=4k+3$ , we can get $(2k+1)(4k+3)(k+1)=\bigg(\frac{b}{2}\bigg)^2$ , same procedure:

$\gcd(2k+1,4k+3)=\gcd(2k+1,1)=1\Bigg|\gcd(2k+1,k+1)=\gcd(-1,k+1)=1\Bigg|\gcd(4k+3,k+1)=\gcd(-1,k+1)=1$ ,

So the integers in the set $\{2k+1,4k+3,k+1\}$ are pairwise coprime, each of them must expressible as square, but this says $\begin{cases}p^2=k+1\\q^2=4k+3\end{cases}$ for positive integers $p,q$ , so $4p^2-q^2=1$ , $(2p-q)(2p+q)=1$ , own solution as $p=\pm\frac{1}{2},q=0$ , contradiction.

As a conclusion, $a(a^2-1)=2b^2$ does not have any positive integer solution, this completes the proof.

that's why answer is 'No'

Expanding the brackets we get:

a³ - a = 2b²

Bringing all the terms to one side:

-a³ + 2b² + a = 0

Factorising:

a (a² + 2b + 1) = 0

Since a ≠ 0 , a² + 2b + 1 must equal 0 in order for the equation to be valid.

a² + 2b + 1 = 0

a² + 2b = -1

As no positive integers could possibly be the values of a and b, the answer is No.

The left side must always be odd and the right side even. No solution in positive integers.

simply put, when you re arrange the terms, you get a^3- a - 2(b^2) =0. This equation is unsolvable, as only has one b term .

By any means it isn't possible by taking the root of both sides ot isn't possible also by multiplying through on the left side bringing the rhs to the left and trying to solve the resulting polynomial it isnt possible as there is no common term I.e an ab term

Among the 3 consecutive integers a-1, a, a+1, only one will be divisible by 3. Thus their product cannot be either a perfect square or twice a perfect square because there will only be one factor of 3.

a(a^2-1)=2b^2 Here a & b are +ive integers If we take a= any integer Then it must to be that the value of 'b' accordance to 'a' is an integer

Now let us take a=2 on puting in given expression we get b^2 = √3 => 'b' is not an integer Hence the given expresion is not true for a&b are integers

Because $a (a^2 -1 ) = 2b^2 ~~~ gcd(a ~,~ a^2 - 1 ) = 1$ ,

we know $a = x^2 ~, ~ (a^2 - 1 ) = 2y^2 ~~ or ~~ a = 2x^2 ~,~ (a^2 - 1 ) = y^2 ~~$

i) $a = x^2 ~, ~ (a^2 - 1 ) = 2y^2 ~$

$x^4 -1 = 2y^2 ~~$

$(x^2 -1)(x^2 + 1) = 2y^2 ~~$

In this case, $gcd( x^2 - 1 ~,~x^2 + 1 ) = 2 ~~$

$x^2 -1 = 2 c^2 , x^2 + 1 = 4d^2 ~~$ or $x^2 -1 = 4 c^2 , x^2 + 1 = 2d^2 ~~$

1) $x^2 -1 = 2 c^2 , x^2 + 1 = 4d^2 ~~$

$1= 4d^2 - x^2 = (2d)^2 - x^2 ~~$ : no solution

2) $x^2 -1 = 4 c^2 ,. x^2 + 1 = 2d^2 ~~$

$1= x^2 - 4c^2 = x^2 - (2c)^2 ~~$ : no solution

so i) case : no solution

ii) $a = 2x^2 ~,~ (a^2 - 1 ) = y^2 ~~$

: $a^2 - y^2 = 1 ~~$ : no solution

By i), ii) we have no solution

At first I tried to make equation and solve by trying to make perfect square etc. but then I tried it like this: $a^2$ -1 = $\frac{b^2}{a}$ . Now we know that LHS is basically a half parabola. and RHS is a half parabola divided by a constant value which in our case is basically square root of LHS ignoring 1. So I don't see a way that both sides can be equal anyway. So I say NO.

The expression on the right has to be even. If a is an even number, the expression on the right will be an even times an odd. If a is an odd number, the expression on the right will be an odd times an even. Therefore the left side cannot be even, so it will never be equal to the right side.

The right hand side of the equation is even. Therefore, the left hand side must also be even. If a is even, the factor in brackets is odd (since the square of an even number is even), and therefore the left hand side in total is odd. If a is odd, then the factor in brackets is even, again resulting in an odd total for the left hand side. Thus, there are no solutions.

Thanks to Sandro for his help : The left hand side is a(a-1)(a+1) and the three factors are coprime, up to factors of 2. In particular, excluding the factors of 2 in the left and right side, we must have that each of a-1, a and a+1 are squares (if c, d, e = f^2, with c,d,e coprime, than c,d,e are all squares). But a-1, a and a+1 can never be simultaneously squares. Even when dividing one of them by 2 (in order to exclude the factor 2), the remaining numbers cannot be squares and distant by 1 or 2.

RHS will be even as 2 is multiplied with an integer.
For LHS,
a is even, LHS is odd
a is odd, LHS is odd
even is not equal to odd
therefore , no solution

since 2b² is even for every b∈ℤ, and a(a²-1) is always odd for every a∈ℤ, they cannot be equal.

Note that $\gcd(a,a^2-1)=1$ , and hence $a$ and $a^2-1$ are coprime. But it is clear that any two coprime factors of $b^2$ whose product is $b^2$ must themselves be squares. So we have $(a,a^2-1)=(2x^2,y^2),(x^2,2y^2)$ , for positive integers $x$ and $y$ .
In the first case, $4x^4-y^2=1$ , which implies $y^2\equiv 3\;\; (\text{mod }4)$ , a contradiction.
In the second case, $x^4-2y^2=1$ , which implies $x$ is odd. So let $x=2m+1$ and factorise, yielding $2m(2m+2)(4m^2+4m+2)=2y^2$ , and thus $4m(m+1)(2m^2+2m+1)=y^2$ . So $m(m+1)(2m^2+2m+1)=m(m+1)(2m(m+1)+1)$ must itself be a perfect square. But $m$ , $m+1$ , and $2m(m+1)+1$ are pairwise coprime, so $m$ , $m+1$ , and $2m(m+1)+1$ must all be perfect squares themselves, implying $m=0$ . This gives $x=1$ , and thus $y=0$ , a contradiction. Hence, there exist no positive integer solutions to the given equation.

Note: it's much faster if you factor $a^2-1=(a-1)(a+1)$ at the start and proceed from the fact that $a-1$ , $a$ , and $a+1$ are pairwise coprime.

a(a² - 1) = 2b².

a and (a² - 1) are coprime.

so a or (a² - 1) must divide 2, and then other must divide b².

case 1:

if a|2 than a = 2.

meaning (2² - 1) = 3. 3 has no natural square root.

case 2:

There is no a such that (a² - 1) = 2.

So there is no solution.